Chapter 2Properties of Triangles
The triangle is the figure that we have spent the most time investigating in our study of geometry through junior high school. Congruence and similarity, parallel lines and angles, the Pythagorean theorem—each of these tools has been used in combination within the single picture of a triangle. In this chapter we will see that the triangle in fact has a much deeper structure. Specifically, we will confirm that a single triangle has several “points of special significance,” and that these points can be constructed rigorously using only the tools learned in junior high school (similarity, the midpoint theorem, and the Pythagorean theorem).
First, in Section 2.1, we examine the ratio in which an angle bisector divides the opposite side. This is the first application of the idea of “dividing a segment in a given ratio”—internal division and external division—and it will serve as the foundation when we later learn the coordinate formulas for internal and external division points in Chapter 4 (Coordinate Geometry) and Chapter 7 (Vectors). In Sections 2.2 and 2.3, we then construct, one by one, working through the steps ourselves, four special points associated with a triangle—the centroid, the orthocenter, the circumcenter, and the incenter—using the midpoint theorem, the Pythagorean theorem, and the idea that “the locus of equidistant points is the perpendicular bisector (the angle bisector).” These are the four principal members of the group of points that have long been known and cherished as the “five centers of a triangle.” Finally, in Section 2.4, we study Ceva's theorem, which decides from ratios alone whether three lines meet at a single point, and Menelaus's theorem, which decides from ratios alone whether three points lie on a single line. At first glance both look like formulas to be memorized, but in fact you can prove them yourself by applying the “parallel lines and ratios” technique used in Section 2.1 exactly as it is—in this chapter we also give those proofs, which are not written in the original notes, as an extension.
- The meaning of internal and external division points of a segment, and the ratio ($AB:AC$) in which the angle bisector and the exterior angle bisector divide the opposite side (or its extension)
- The fact that the centroid of a triangle divides each median internally in the ratio $2:1$ (a proof using the midpoint theorem), and the proof of the median theorem (Apollonius's theorem) $AB^2+AC^2=2(AM^2+BM^2)$
- The definition of the orthocenter (the intersection of the perpendiculars from the vertices to the opposite sides) and a proof that the three perpendiculars meet at a single point
- The definitions and proofs of existence of the circumcenter (the intersection of the perpendicular bisectors of the sides; the center of the circumcircle) and the incenter (the intersection of the bisectors of the angles; the center of the incircle)
- Ceva's theorem and Menelaus's theorem and their proofs, and why the mnemonic for remembering them takes the form it does
Source notes: Yasuhide Mochizuki, Mathematics Notes: High School Mathematics (handwritten lecture notes, in Japanese), pp. 3–6.
2.1 Angle Bisectors and Internal and External Division
If we draw from vertex $A$ of triangle $ABC$ the ray that bisects $\angle A$, it meets the opposite side $BC$ at some point. This point is not necessarily the exact midpoint of $BC$—if the lengths of $AB$ and $AC$ differ, the intersection is shifted away from the middle. To state precisely how far it is shifted, let us first define the phrase “dividing a segment in a ratio.”
Definition 2.1: Internal and external division points of a segment
Suppose that a point $P$ lies on the segment $BC$ joining two points $B,C$, and that, for positive numbers $m,n$,
$$ BP:PC=m:n $$holds. Then we say that $P$ internally divides the segment $BC$ in the ratio $m:n$, and we call $P$ an internal division point of $BC$. On the other hand, suppose that a point $Q$ lies on the line $BC$ outside the segment $BC$ and satisfies (where $m\neq n$)
$$ BQ:CQ=m:n $$Then we say that $Q$ externally divides the segment $BC$ in the ratio $m:n$, and we call $Q$ an external division point of $BC$.
There is a reason for attaching the condition “$m\neq n$” to the definition of the external division point. If $m=n$, then $BQ=CQ$, and the only point satisfying this is the midpoint of $BC$—but the midpoint is a point inside the segment $BC$, not outside it. In other words, when $m=n$, no external division point exists in the first place. This fact will reappear in Section 2.1.2, in the discussion of the exterior angle bisector.
2.1.1 The interior angle bisector
First, let us consider the bisector of $\angle A$ itself (the interior angle).
Theorem 2.1: The interior angle bisector and the internal division point
In $\triangle ABC$, if the bisector of $\angle A$ meets side $BC$ at $P$, then $P$ divides side $BC$ internally so that
$$ BP:PC=AB:AC $$holds.
The strategy of the proof is to “draw one auxiliary parallel line and replace the two equal angles formed by the bisector with the equal angles formed by the parallel line.” This technique is the central tool of this chapter; we will use it repeatedly later, for the exterior angle bisector, Ceva's theorem, and Menelaus's theorem.
Proof (Theorem 2.1)
Extend side $BA$ beyond $A$, and on this extension take the point $D$ where it meets the line through $C$ parallel to line $AP$ (Figure 2.1). In other words, we define $D$ so that $AP\parallel CD$.
Line $BD$ ($=$ line $BA$ extended) is a transversal cutting the two parallel lines $AP,CD$. Therefore the corresponding angles are equal, and
$$ \angle BAP=\angle ADC $$holds. Also, line $AC$ is a transversal cutting $AP,CD$ as well, so the alternate angles are equal, and
$$ \angle PAC=\angle ACD $$holds. Meanwhile, since $AP$ is the bisector of $\angle A$, we have $\angle BAP=\angle PAC$. Putting these three equalities side by side, we get
$$ \angle ADC=\angle BAP=\angle PAC=\angle ACD $$so in particular $\angle ADC=\angle ACD$. In $\triangle ACD$ the two angles $\angle ADC,\angle ACD$ are equal, so the sides $AC,AD$ opposite these angles, respectively, are also equal (“a triangle with two equal angles is isosceles”—the converse of “the base angles of an isosceles triangle are equal,” a theorem learned in junior high school):
$$ AC=AD \qquad \cdots\text{①} $$Next we use $AP\parallel CD$. $\angle ABP=\angle DBC$ because they are the same angle ($P$ lies in the direction of $C$ as seen from $B$, and $A$ lies in the direction of $D$ as seen from $B$, so $\angle ABP$ and $\angle DBC$ refer to exactly the same angle). Also, since line $BD$ cuts $AP,CD$, for the same reason as before $\angle BAP=\angle BDC$ ($=\angle ADC$). Since two pairs of angles are respectively equal,
$$ \triangle ABP \backsim \triangle DBC $$($\backsim$ is the symbol for “is similar to”). In similar figures the ratios of corresponding sides are all equal, so
$$ BA:BD=BP:BC \qquad (\ast) $$holds. Here, since $D$ is a point on the extension beyond $A$, $BD=BA+AD$, and similarly, since $P$ is a point between the points $B,C$, $BC=BP+PC$. We substitute these into $(\ast)$ and simplify (filling in the steps between the lines). Since $(\ast)$ can be written $BA\cdot BC=BP\cdot BD$, we obtain
$$ BA(BP+PC)=BP(BA+AD) $$ $$ BA\cdot BP+BA\cdot PC=BP\cdot BA+BP\cdot AD $$Subtracting the common term $BA\cdot BP$ ($=BP\cdot BA$) from both sides gives
$$ BA\cdot PC=BP\cdot AD $$that is,
$$ BA:AD=BP:PC \qquad \cdots\text{②} $$Finally, substituting ① ($AC=AD$) for $AD$ in ② gives
$$ AB:AC=BP:PC $$and the theorem is proved.
∎
Example 2.1: Using the interior angle bisector
In $\triangle ABC$, suppose $AB=6$, $AC=4$, and $BC=5$. Let $P$ be the point where the bisector of $\angle A$ meets side $BC$. Find the lengths of the segments $BP,PC$.
Solution. By Theorem 2.1, $BP:PC=AB:AC=6:4=3:2$. We have $BP+PC=BC=5$, and $BP:PC=3:2$ means “of the $3+2=5$ equal parts into which $BC$ is divided, $3$ parts and $2$ parts,” so
$$ BP=5\times\frac{3}{5}=3,\qquad PC=5\times\frac{2}{5}=2 $$(Check: $BP+PC=3+2=5=BC$ and $BP:PC=3:2=6:4=AB:AC$, so both conditions are satisfied.)
2.1.2 The exterior angle bisector
In a $\triangle ABC$ with $AB\neq AC$, now draw the ray that bisects the exterior angle of $\angle A$ (the one that is not the adjacent interior angle, on the side of $180^\circ-\angle A$). This ray does not meet side $BC$ itself, but it does meet the extension of side $BC$.
Theorem 2.2: The exterior angle bisector and the external division point
In a $\triangle ABC$ with $AB\neq AC$, if the bisector of the exterior angle of $\angle A$ meets the extension of side $BC$ at $Q$, then $Q$ divides side $BC$ externally so that
$$ BQ:CQ=AB:AC $$holds.
Proof (Theorem 2.2)
Below we prove the case $AB\gt AC$ (the Math Note after the proof supplements how the positions of $E,Q$ change when $AB\lt AC$).
Extend side $BA$ beyond $A$ and take a point $D$ on the extension ($\angle DAQ$ is the exterior angle, that is, the angle that is bisected). Also draw the line through $C$ parallel to line $AQ$, and let $E$ be its intersection with side $AB$ (the segment $BA$ itself). In other words, we define $E$ so that $EC\parallel AQ$ (Figure 2.2).
Line $BD$ ($=$ the extension of line $AB$) is a transversal cutting the two parallel lines $EC,AQ$, so the corresponding angles are equal, and
$$ \angle DAQ=\angle AEC $$holds. Also, line $AC$ is a transversal cutting $EC,AQ$ as well, so the alternate angles are equal, and
$$ \angle QAC=\angle ACE $$holds. Since $AQ$ is the bisector of the exterior angle $\angle DAC$, we have $\angle DAQ=\angle QAC$. Therefore
$$ \angle AEC=\angle DAQ=\angle QAC=\angle ACE $$so in particular $\angle AEC=\angle ACE$. Since these two angles of $\triangle AEC$ are equal, the sides $AC,AE$ opposite them are also equal:
$$ AE=AC \qquad \cdots\text{①} $$Next, $\angle EBC=\angle ABQ$ because they are the same angle ($E$ is a point in the direction of $A$ as seen from $B$, and $Q$ is a point in the direction of $C$ as seen from $B$). Also, from $EC\parallel AQ$ and the transversal $BD$ we get $\angle BEC=\angle BAQ$ (corresponding angles: the direction of $B$ as seen from $E$ and the direction of $B$ as seen from $A$ are on the same side of the same line, so they correspond). Since two pairs of angles are respectively equal,
$$ \triangle EBC \backsim \triangle ABQ $$and from the ratios of corresponding sides
$$ EB:AB=BC:BQ \qquad (\ast\ast) $$holds. Here $EA=AB-EB$ ($E$ is a point between the points $B,A$) and $CQ=BQ-BC$ ($C$ is a point between the points $B,Q$, and $Q$ lies on the extension beyond $C$). Since $(\ast\ast)$ can be written $EB\cdot BQ=AB\cdot BC$, we have
$$ BA\cdot CQ=AB(BQ-BC)=AB\cdot BQ-AB\cdot BC $$ $$ EA\cdot BQ=(AB-EB)BQ=AB\cdot BQ-EB\cdot BQ=AB\cdot BQ-AB\cdot BC $$The right-hand sides coincide, so $BA\cdot CQ=EA\cdot BQ$, that is,
$$ BA:EA=BQ:CQ \qquad \cdots\text{②} $$is obtained. Substituting ① ($AE=AC$) for $EA$ in ② gives
$$ AB:AC=BQ:CQ $$and the theorem is proved.
∎
Caution: Why is $AB\neq AC$ necessary?
Theorem 2.2 assumed $AB\neq AC$. If $AB=AC$ (an isosceles triangle), then the bisector of the exterior angle of $\angle A$ is parallel to the base $BC$, and the intersection point $Q$ does not exist. Indeed, recall that in Definition 2.1 we required $m\neq n$ for an external division point—if we forcibly write down the conclusion of Theorem 2.2 when $AB=AC$, we get $BQ:CQ=AB:AC=1:1$, but this is a ratio that does not exist as the ratio of an external division point. In the language of figures, this says that “the bisector of the exterior angle at the apex of an isosceles triangle coincides with the perpendicular bisector of the base, and moves away parallel to the base without ever meeting it.”
Math Note: How do the positions of $E,Q$ change when $AB\lt AC$?
In the proof we treated the case $AB\gt AC$. When $AB\lt AC$ (that is, when $AC$ is the longer side), exactly the same argument can be used as it stands, simply by interchanging the roles of $B$ and $C$ throughout the proof. Concretely, let $D'$ be the point on the extension of side $CA$ beyond $A$, and draw the line through the point $B$ parallel to the line $AQ$, letting $E'$ be its intersection with side $AC$ (the segment $CA$ itself). If we merely read $B,E,D,C$ in the proof as $C,E',D',B$ respectively, every argument using corresponding angles, alternate angles, and similarity goes through unchanged: first we obtain $AE'=AB$ (the equation corresponding to ① “$AE=AC$” in the proof), and then $AC:AB=CQ:BQ$ (the relation corresponding to ② in the proof, rearranged). This is the same equation as $AB:AC=BQ:CQ$ with only the order of the terms on both sides interchanged, so the conclusion does not change. What differs is that the external division point $Q$ now appears not on the side of $C$ but on the extension beyond $B$.
In general, the structure of this proof shows the following: the external division point $Q$ appears on the extension beyond the vertex (other than $A$) corresponding to the shorter of the sides $AB,AC$. In Example 2.2 ($AB=5\gt AC=3$), Q appears on the extension beyond the vertex $C$ of the shorter side $AC$, and the rule of thumb used in its solution (“the shorter the side on the $AC$ side, the farther the exterior angle bisector runs off toward $C$”) is nothing other than a restatement of this fact. In Exercise 2.1 ($AB=5\lt AC=8$), conversely, $Q$ appears on the extension beyond the vertex $B$ of the shorter side $AB$.
Example 2.2: Using the exterior angle bisector
In $\triangle ABC$, suppose $AB=5$, $AC=3$, and $BC=4$. Let $Q$ be the intersection of the bisector of the exterior angle of $\angle A$ with the extension of side $BC$. Find the lengths of the segments $BQ,CQ$.
Solution. Since $AB\neq AC$ ($5\neq3$), Theorem 2.2 applies, and $BQ:CQ=AB:AC=5:3$. Since $AB\gt AC$, the point $Q$ lies on the extension beyond $C$ (the shorter the side on the $AC$ side, the farther the exterior angle bisector runs off toward $C$). Let $CQ=t$. Then $BQ=BC+CQ=4+t$, and
$$ \frac{BQ}{CQ}=\frac{4+t}{t}=\frac{5}{3} $$gives $3(4+t)=5t$, that is, $12+3t=5t$, $2t=12$, and $t=6$. Hence
$$ CQ=6,\qquad BQ=4+6=10 $$(Check: $BQ:CQ=10:6=5:3=AB:AC$, so the condition is satisfied.)
2.2 The Five Centers of a Triangle (1) — Centroid, Median Theorem, and Orthocenter
From here on, we look for the “special points” hidden inside a single triangle. The first one we treat is the point formed when we join every vertex to “the exact middle of the opposite side.”
Definition 2.2: Median and centroid
A segment joining a vertex of a triangle to the midpoint of its opposite side is called the median from that vertex. A triangle has three vertices, so it has three medians, but, as we confirm in Theorem 2.3, these three medians always meet at exactly one point. This point of intersection is called the centroid of the triangle.
2.2.1 The centroid divides each median internally in the ratio 2:1
Theorem 2.3: The centroid and the ratio on the medians
Let $L,M$ be the midpoints of the sides $BC,CA$ of $\triangle ABC$, respectively. If $G$ is the intersection of the medians $AL$ and $BM$, then
$$ AG:GL=BG:GM=2:1 $$holds. Moreover, if we let $N$ be the midpoint of side $AB$ and draw the third median $CN$, it also passes through the same point $G$.
Proof (Theorem 2.3)
Since $M$ is the midpoint of side $CA$ and $L$ is the midpoint of side $CB$, applying the midpoint theorem that we learned in junior high school mathematics (the theorem that the segment joining the midpoints of two sides is parallel to the remaining side and half as long) to $\triangle CAB$, we find that
$$ ML\parallel AB, \qquad ML=\frac{1}{2}AB \qquad \cdots\text{①} $$holds (Figure 2.3).
Let $G$ be the intersection of the medians $AL$ and $BM$. Because $ML\parallel AB$, the line $AL$ (the line through $A,G,L$, not the line through $M,L$) is a transversal cutting the two lines $AB,ML$, and the alternate angles are equal, so $\angle GAB=\angle GLM$. Similarly, the line $BM$ is a transversal, and $\angle GBA=\angle GML$. Moreover, $\angle AGB$ and $\angle LGM$, formed where the two lines $AL,BM$ meet at $G$, are vertical angles (angles facing each other), so they are equal. Since the three pairs of angles are equal (two pairs would be enough, but we have checked all three to be safe), we have
$$ \triangle GAB \backsim \triangle GLM $$The similarity ratio is $AB:LM=2:1$ by ①, so all the ratios of corresponding sides are also $2:1$:
$$ AG:GL=BG:GM=AB:LM=2:1 $$This shows that the medians $AL,BM$ are each divided internally at $G$ in the ratio $2:1$.
Verification that the third median passes through the same point. Consider also the medians $AL$ and $CN$ ($N$ is the midpoint of $AB$). If we carry out exactly the same argument with $\triangle BCA$ in place of $\triangle CAB$ (apply the midpoint theorem to $N,L$—the midpoints of $AB,CB$—to obtain $NL\parallel CA$ and $NL=\frac12 CA$, and then construct $\triangle G'AC\backsim\triangle G'LN$), then, letting $G'$ be the intersection of the medians $AL$ and $CN$, we again find that $AG':G'L=2:1$. However, on the median $AL$ there is only one point lying “at the $2:1$ position from $A$” (the point dividing a segment internally in the ratio $2:1$ is uniquely determined). Therefore $G$ and $G'$ are the same point. That is, the median $CN$ also passes through the point $G$. Hence the three medians $AL,BM,CN$ all meet at the same point $G$, and $G$ divides every median internally in the ratio $2:1$ from the vertex side.
∎
Math Note: Why is this last paragraph necessary?
Merely showing that “two medians meet in the ratio $2:1$” does not, in fact, tell us that all three medians pass through a single point. Only after checking that the third median really passes through the same point can we safely use the name “centroid” for a single point. Even the fact that “all three meet at one point,” which seems obvious at first glance, is a theorem that needs a proper proof.
Example 2.3: Using the ratio of the centroid
In $\triangle ABC$, the median $AL$ drawn from the vertex $A$ has length $9$. For the centroid $G$, find the lengths $AG,GL$.
Solution. By Theorem 2.3, $AG:GL=2:1$, and $AG+GL=AL=9$. Since $2:1$ means “$2$ parts and $1$ part out of $9$ divided into $2+1=3$ equal parts,” we have
$$ AG=9\times\frac{2}{3}=6,\qquad GL=9\times\frac{1}{3}=3 $$(Check: $AG+GL=6+3=9=AL$ and $AG:GL=6:3=2:1$, so the condition is satisfied.)
2.2.2 The median theorem
Next, we prove an equation that connects the length of a median itself to the lengths of the three sides of the triangle.
Theorem 2.4: The median theorem
Let $M$ be the midpoint of side $BC$ of $\triangle ABC$. Then
$$ AB^2+AC^2=2\bigl(AM^2+BM^2\bigr) $$holds.
Proof (Theorem 2.4)
Drop the perpendicular from the vertex $A$ to side $BC$, and let $H$ be its foot (we give the proof for the case where $H$ lies between $M$ and $C$, as in Figure 2.4; the treatment of the other positions of $H$ is supplied in the note that follows). Since $\triangle ABH,\triangle ACH$ are both right triangles with the right angle at $H$, the Pythagorean theorem gives
$$ AB^2=BH^2+AH^2, \qquad AC^2=CH^2+AH^2 $$From the configuration in the figure, $BH=BM+MH$ and $CH=MC-MH$ (since $H$ lies between the points $M,C$, the distance from $C$ to $H$ is $MC$ minus $MH$). Substituting these and adding the two equations, we get
$$ AB^2+AC^2=(BM+MH)^2+AH^2+(MC-MH)^2+AH^2 $$Expand each of the two squares on the right-hand side ($(x+y)^2=x^2+2xy+y^2$, $(x-y)^2=x^2-2xy+y^2$):
$$ =BM^2+2\,BM\!\cdot\!MH+MH^2+AH^2+MC^2-2\,MC\!\cdot\!MH+MH^2+AH^2 $$Here, since $M$ is the midpoint of $BC$, we have $BM=MC$. Therefore $2\,BM\!\cdot\!MH$ and $-2\,MC\!\cdot\!MH$ have the same magnitude and opposite signs because $BM=MC$, so they cancel. Also, $BM^2+MC^2=2BM^2$ (because $BM=MC$), the two $MH^2$ terms give $2MH^2$, and the two $AH^2$ terms give $2AH^2$. Collecting terms, we obtain
$$ AB^2+AC^2=2BM^2+2MH^2+2AH^2=2\bigl(BM^2+MH^2+AH^2\bigr) $$Furthermore, $\triangle AMH$ is also a right triangle with the right angle at $H$, so the Pythagorean theorem gives $MH^2+AH^2=AM^2$. Substituting this, we get
$$ AB^2+AC^2=2\bigl(BM^2+AM^2\bigr)=2\bigl(AM^2+BM^2\bigr) $$which proves the theorem.
∎
Math Note: What happens when $H$ is not between $M$ and $C$?
In Figure 2.4 we chose a triangle in which the foot $H$ of the perpendicular happens to lie between the points $M,C$, but depending on the shape of $\triangle ABC$, $H$ may lie between the points $B,M$, may coincide with $M$ itself (the case of an isosceles triangle with $AB=AC$), or may even fall outside the triangle (the case of an obtuse triangle). In every case, if we rewrite $BH,CH$ as sums or differences of $BM,MH$, paying attention to signs (for example, if $H$ is between the points $B,M$, then $BH=BM-MH$ and $CH=MC+MH$), the same computation reaches the same conclusion $AB^2+AC^2=2(AM^2+BM^2)$. A clearer proof that handles all the cases at once can be obtained by using the law of cosines, which we study in Chapter 3, or vectors, which we study in Chapter 7 (we need only start from $\vec{AM}=\frac12(\vec{AB}+\vec{AC})$).
Example 2.4: Using the median theorem
In $\triangle ABC$, suppose $AB=9$, $AC=7$, and $BC=8$. Letting $M$ be the midpoint of side $BC$, find the length of the median $AM$.
Solution. We have $BM=\dfrac{BC}{2}=4$. By the median theorem (Theorem 2.4),
$$ AB^2+AC^2=2\bigl(AM^2+BM^2\bigr) $$ $$ 9^2+7^2=2\bigl(AM^2+4^2\bigr) $$ $$ 81+49=2\,AM^2+32 $$ $$ 130-32=2\,AM^2 $$ $$ AM^2=\frac{98}{2}=49 $$Since $AM\gt 0$, we have $AM=7$.
2.2.3 The orthocenter
Definition 2.3: Orthocenter
From each vertex of a triangle, drop a perpendicular to the opposite side (or its extension). The point where these three perpendiculars meet is called the orthocenter of the triangle.
Figure 2.5 appears to show three perpendiculars meeting at one point, but, as in the case of the centroid, this is a fact that requires proof. For the proof, we draw three lines, each through a vertex of the original triangle $ABC$ and parallel to the opposite side, and thereby form a larger triangle $A^*B^*C^*$. Here we choose $A^*,B^*,C^*$ so that $A,B,C$ are the midpoints of the sides $B^*C^*,\,C^*A^*,\,A^*B^*$ of $A^*B^*C^*$, respectively (Figure 2.6).
Intuition: Why the three perpendiculars that define the orthocenter meet at one point
$A$ is the midpoint of $B^*C^*$, and moreover, by a generalization of the midpoint theorem, side $BC$ is parallel to side $B^*C^*$ (since $B,C$ are the midpoints of sides $A^*C^*,A^*B^*$, respectively, $BC$ is a midsegment of $\triangle A^*B^*C^*$). Therefore “a line perpendicular to $BC$” and “a line perpendicular to $B^*C^*$” point in the same direction. That is, the perpendicular drawn from vertex $A$ of $\triangle ABC$ to side $BC$ is the line through the midpoint $A$ of $B^*C^*$ perpendicular to $B^*C^*$; in other words, it is the perpendicular bisector of side $B^*C^*$ itself. Carrying out the same argument for $B,C$, we see that the three perpendiculars of $\triangle ABC$ coincide exactly with the three perpendicular bisectors of $\triangle A^*B^*C^*$. Now, as we prove in Theorem 2.5 of Section 2.3, in any triangle the three perpendicular bisectors always meet at one point (the circumcenter). Therefore the three perpendiculars of $\triangle ABC$ also always meet at one point—this point is the orthocenter.
Example 2.5: The orthocenter of a right triangle
Show that if $\angle C=90^\circ$ in $\triangle ABC$, then the orthocenter is the vertex $C$ itself.
Solution. Since $\angle C=90^\circ$, side $CA$ and side $CB$ are perpendicular. Therefore side $CB$ itself plays the role of “the perpendicular drawn from vertex $B$ to the opposite side $CA$”—this is because $CB$ passes through the point $B$ and $CB\perp CA$, so the line $CB$ is none other than “the line through $B$ perpendicular to $CA$.” Similarly, the line $CA$ coincides with “the line through $A$ perpendicular to the opposite side $CB$.” That is, the perpendicular from vertex $B$ is the line $CB$ itself, and the perpendicular from vertex $A$ is the line $CA$ itself, and these two lines obviously meet at the point $C$. Since the orthocenter is the intersection point of all three perpendiculars, the orthocenter is the point $C$.
2.3 The Five Centers of a Triangle (2) — Circumcenter, Incenter, and Centroid
In the previous section we constructed the centroid and the orthocenter by using “the midpoint of the opposite side” and “the foot of a perpendicular.” In this section, from the viewpoint of “the set of equidistant points,” we construct two more special points—the circumcenter and the incenter. The key to the approach is the following simple fact: all points that are equidistant from two points $X,Y$ lie on the perpendicular bisector of the segment $XY$ (and, conversely, every point on the perpendicular bisector is necessarily equidistant from the points $X,Y$). Similarly, for angle bisectors: all points that are equidistant from the two sides of an angle (the two lines forming the angle) lie on the bisector of that angle. Using these two facts, the existence of the circumcenter and the incenter becomes a “theorem that can be proved.”
2.3.1 The circumcenter
Definition 2.4: Circumcenter
The point of intersection of the perpendicular bisectors of the sides of a triangle is called the circumcenter of that triangle.
Theorem 2.5: Existence of the circumcenter and the circumcircle
For any triangle $ABC$, the perpendicular bisectors of the three sides meet at exactly one point $O$. This point $O$ is equidistant from the three vertices $A,B,C$, and we can draw a circle centered at $O$ passing through $A,B,C$ (this circle is called the circumcircle, and its radius is called the circumradius).
Proof (Theorem 2.5)
The perpendicular bisector of side $AB$ and the perpendicular bisector of side $BC$ are not parallel (as long as $A,B,C$ do not lie on a single line), so they meet at some point $O$. Since $O$ lies on the perpendicular bisector of side $AB$, by the fact stated at the beginning we have
$$ OA=OB $$Similarly, since $O$ lies on the perpendicular bisector of side $BC$, we have
$$ OB=OC $$Combining these two, $OA=OB=OC$; that is, $O$ is equidistant from all three vertices. In particular, $OA=OC$ holds, and this is exactly the statement that “$O$ lies on the perpendicular bisector of side $CA$” (the converse direction of the fact stated at the beginning). Therefore, the perpendicular bisector of side $CA$ also passes through the point $O$. In sum, all three perpendicular bisectors pass through the point $O$, so $O$ is determined as a single point. If we draw a circle centered at $O$ with the common length $OA=OB=OC$ as its radius, that circle passes through all three points $A,B,C$.
∎
Point: The circumcircle encloses the triangle from the outside
The circumcircle passes through the three vertices $A,B,C$ and contains the whole triangle $ABC$ in its interior (it does not touch the sides anywhere between the vertices; it passes through the vertices only). It is helpful to remember this with the image that “the triangle fits snugly inside the circumcircle.” In contrast, the incircle that we study next is, conversely, a circle that fits inside the triangle while touching each side.
Example 2.6: The circumcenter of a right triangle
In $\triangle ABC$ with $\angle A=90^\circ$, show that the circumcenter is the midpoint of side $BC$ (the hypotenuse).
Solution. Let $M$ be the midpoint of side $BC$. Since $\angle A=90^\circ$, the Pythagorean theorem (applied to $\triangle ABC$ itself) gives
$$ AB^2+AC^2=BC^2 $$On the other hand, applying the median theorem (Theorem 2.4) to $\triangle ABC$ gives
$$ AB^2+AC^2=2\bigl(AM^2+BM^2\bigr) $$Since $M$ is the midpoint of $BC$, $BM=\dfrac{BC}{2}$; that is, $BM^2=\dfrac{BC^2}{4}$. The left-hand sides of the two equations are the same, $AB^2+AC^2$, so the right-hand sides are also equal:
$$ BC^2=2AM^2+2BM^2=2AM^2+\frac{BC^2}{2} $$Rearranging gives $BC^2-\dfrac{BC^2}{2}=2AM^2$, that is, $\dfrac{BC^2}{2}=2AM^2$, and hence $AM^2=\dfrac{BC^2}{4}=BM^2$. Since lengths are positive, $AM=BM$. Since $M$ was the midpoint of $BC$ to begin with, we also have $BM=CM$. Therefore
$$ AM=BM=CM $$and $M$ is a point equidistant from the three vertices $A,B,C$. As we saw in the proof of Theorem 2.5, there is only one point equidistant from the three vertices (the circumcenter), so $M$ itself is the circumcenter.
2.3.2 The incenter
Definition 2.5: Incenter
The point of intersection of the bisectors of the angles of a triangle is called the incenter of that triangle.
Theorem 2.6: Existence of the incenter and the incircle
For any triangle $ABC$, the bisectors of the three angles meet at exactly one point $I$. This point $I$ is equidistant from the three sides (more precisely, from the lines containing them), and if we draw the circle centered at $I$ whose radius is that distance, it is tangent to all three sides (this circle is called the incircle).
Proof (Theorem 2.6)
The bisector of $\angle A$ and the bisector of $\angle B$ meet at exactly one point $I$ inside the triangle. Since $I$ lies on the bisector of $\angle A$, by the fact stated at the beginning (angle bisectors and equal distances), the distance from $I$ to the line containing side $AB$ equals the distance to the line containing side $AC$. Similarly, since $I$ lies on the bisector of $\angle B$, the distance from $I$ to side $BA$ ($=AB$) and the distance to side $BC$ are also equal. Since “the distance to side $AB$” is common to these two facts,
$$ (\text{distance from $I$ to side $CA$})=(\text{distance from $I$ to side $AB$})=(\text{distance from $I$ to side $BC$}) $$and so $I$ is equidistant from all three sides. In particular, “the distance to side $CA$ equals the distance to side $BC$” holds, and this is exactly the statement that “$I$ lies on the bisector of $\angle C$.” Therefore, the bisector of $\angle C$ also passes through the point $I$, and the three angle bisectors all meet at $I$. Let $D,E,F$ be the feet of the perpendiculars to sides $BC,CA,AB$, respectively. Then $ID=IE=IF$ (the common distance to the three sides), so the circle with center $I$ and radius $ID$ is tangent to all three sides (at the feet of the perpendiculars $D,E,F$).
∎
Point: The incircle touches each side of the triangle from the inside
The incircle is a circle that fits snugly inside the triangle while touching (being tangent to) each of the three sides at one point apiece. Whereas the circumcircle “passes through the vertices and contains the triangle in its interior,” the incircle “touches the sides and fits inside the triangle”—exactly the opposite relationship.
Math Note: A formula for the inradius in terms of the area
Using the incenter $I$ and the area $S=[\triangle ABC]$, we can build a formula for the inradius $r=ID=IE=IF$. Joining $I$ to the three vertices divides $\triangle ABC$ into three smaller triangles, $\triangle IBC,\triangle ICA,\triangle IAB$. The area of each can be written as “base (a side of the triangle) $\times$ height $(=r)\div2$,” so with $BC=a,CA=b,AB=c$ we have
$$ S=\frac12 r\,a+\frac12 r\,b+\frac12 r\,c=\frac12 r(a+b+c) $$Writing the semiperimeter of the triangle as $s=\dfrac{a+b+c}{2}$ ($s$ is the initial of “semiperimeter”), this equation becomes $S=rs$, that is,
$$ r=\frac{S}{s} $$Let us actually use this in the next example.
Example 2.7: The inradius
Find the inradius $r$ of the right triangle whose sides have lengths $3,4,5$ (let $\angle C=90^\circ$, $CA=3,CB=4,AB=5$).
Solution. Since $\angle C=90^\circ$, the area is $S=\dfrac12\times3\times4=6$. The semiperimeter is $s=\dfrac{3+4+5}{2}=6$. Using the formula $r=\dfrac{S}{s}$ from the Math Note, we obtain
$$ r=\frac{6}{6}=1 $$Thus the inradius is 1.
2.3.3 Summary of the centroid—its physical meaning
In Section 2.2, we constructed the centroid of a triangle as “the intersection of the three medians.” If we draw it in the same figure as the circumcenter and the incenter, the centroid $G$ appears as follows (Figure 2.9).
Point: The centroid is the “center of weight” in the physical sense
If we prepare a triangular plate made of a uniform material (a plate of uniform thickness and density), the plate can be supported and balanced perfectly horizontally precisely when the support is placed directly below (or directly above) the centroid. The Japanese name for the centroid, which literally means “center of weight,” comes from this physical property. The same idea is also used in materials science when we consider the mechanical balance of a thin, plate-shaped specimen.
Errors and inconsistent notation in the original notes
The original notes contain two inconsistencies in how points are named in this section. (1) In the explanatory text introducing the incenter, the center is called “$I$” ($ID=IE=IF$), but in the figure the center is labeled “$Q$.” (2) Of the two figures dealing with the centroid, in the figure used to prove the ratio of the medians (corresponding to Figure 2.3 in this book) the point of intersection is labeled “$G$,” whereas in the figure defining the centroid (corresponding to Figure 2.9 in this book) the point of intersection is labeled “$R$.” In both cases the same point is called by different names; in this book we have standardized on $I$ for the incenter and $G$ for the centroid throughout.
Math Note: The one of the five centers not covered in this chapter
Together, the centroid, the orthocenter, the circumcenter, and the incenter are often called “the five centers of a triangle” (the title of this section follows this usage). In fact, however, there are five points that are typically called the five centers, and they are the four we have learned in this chapter plus the excenter. An excenter is a point constructed as the intersection of the bisector of one interior angle and the bisectors of the exterior angles at the remaining two vertices, and a triangle has three excenters. If you regard this construction as a direct application of the idea of the “exterior angle bisector” learned in Section 2.1.2 to the argument about the intersection of the angle bisectors (the incenter), the tools for the proof are already in place. As an advanced topic, if you are interested, try drawing a figure yourself and checking it.
Digression: The Euler line
If we draw the circumcenter $O$ and centroid $G$ constructed in this section, together with the orthocenter $H$ constructed in Section 2.2, in a single triangle at the same time, then in fact the three points $O,G,H$ always lie on one line, and moreover $G$ divides the segment $OH$ internally in the ratio $OG:GH=1:2$. This line is called the Euler line, named after its discoverer. Figure 2.10 is a check, carried out for the triangle we have been using throughout this section, in which $O,G,H$ are actually computed and found to lie on one line. A concise proof of this fact is most transparent when vectors are used (after learning about vectors in Chapter 7, one can verify it by starting from the relation $\vec{OA}+\vec{OB}+\vec{OC}=\vec{OH}$).
2.4 Ceva's Theorem and Menelaus's Theorem
Finally, we study two theorems that, when three points are taken on the sides of a triangle (or on their extensions), let us decide from a product of ratios alone whether “the three lines drawn from them meet at one point” and whether “those three points lie on one line.” In the original notes the proofs are omitted (they say only “proof omitted”), but in this book we also supply proofs, applying the techniques used in Section 2.1 (similar triangles and ratios of segments cut by parallel lines).
2.4.1 Ceva's theorem
Theorem 2.7: Ceva's theorem
On the sides $AB,BC,CA$ of $\triangle ABC$ (or on their extensions), take points $R,P,Q$, respectively. If the three segments (or their extensions) $AP,BQ,CR$ meet at one point, then
$$ \frac{AR}{RB}\times\frac{BP}{PC}\times\frac{CQ}{QA}=1 $$holds.
Proof (Theorem 2.7)
Suppose that $AP,BQ,CR$ meet at a point $X$ (Figure 2.11). We use the fact that “the ratio of the areas of two triangles that share a vertex and whose bases lie on the same line equals the ratio of the lengths of their bases” (this holds because the area of a triangle $=\frac12\times$ base $\times$ height, and the height from the common vertex to the base is the same).
First, looking at the base $BC$ from vertex $A$ and from vertex $X$, respectively, we have
$$ \frac{BP}{PC}=\frac{[\triangle ABP]}{[\triangle ACP]}=\frac{[\triangle XBP]}{[\triangle XCP]} $$(Here $[\ \cdot\ ]$ denotes area. The first area ratio is the one obtained by looking at $B,P,C$ on the base $BC$ from vertex $A$, and the second is the one obtained from vertex $X$.) In general, if $\dfrac{a}{b}=\dfrac{c}{d}$, then $\dfrac{a}{b}=\dfrac{a-c}{b-d}$ holds (if we let $a=kb,c=kd$, it follows immediately from $a-c=k(b-d)$), so
$$ \frac{BP}{PC}=\frac{[\triangle ABP]-[\triangle XBP]}{[\triangle ACP]-[\triangle XCP]} $$Now, since $X$ is a point on segment $AP$, segment $BX$ divides $\triangle ABP$ into $\triangle ABX$ and $\triangle XBP$: $[\triangle ABP]=[\triangle ABX]+[\triangle XBP]$, that is, $[\triangle ABP]-[\triangle XBP]=[\triangle ABX]$. Similarly, segment $CX$ divides $\triangle ACP$ into $\triangle ACX$ and $\triangle XCP$, so $[\triangle ACP]-[\triangle XCP]=[\triangle ACX]$. Hence
$$ \frac{BP}{PC}=\frac{[\triangle ABX]}{[\triangle ACX]} \qquad \cdots\text{①} $$Carrying out exactly the same argument for vertex $B$ and the point $Q$ on side $CA$, and for vertex $C$ and the point $R$ on side $AB$ (using that $X$ lies on each of $BQ,CR$), we obtain
$$ \frac{CQ}{QA}=\frac{[\triangle BCX]}{[\triangle BAX]} \qquad \cdots\text{②}, \qquad\qquad \frac{AR}{RB}=\frac{[\triangle CAX]}{[\triangle CBX]} \qquad \cdots\text{③} $$If we multiply together all the right-hand sides of ①, ②, ③, then, since $[\triangle ABX]=[\triangle BAX]$, $[\triangle ACX]=[\triangle CAX]$, $[\triangle BCX]=[\triangle CBX]$ (triangles with the same three vertices have equal areas regardless of the order in which the vertices are listed), the numerator and the denominator become products of exactly the same three areas, and they cancel neatly:
$$ \frac{BP}{PC}\times\frac{CQ}{QA}\times\frac{AR}{RB}=\frac{[\triangle ABX]}{[\triangle ACX]}\times\frac{[\triangle BCX]}{[\triangle BAX]}\times\frac{[\triangle CAX]}{[\triangle CBX]}=1 $$Therefore $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$.
∎
Math Note: The converse also holds
In fact, the converse of Theorem 2.7—“if $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$ holds, then $AP,BQ,CR$ meet at one point”—is also true. We state only the strategy of the proof: let $X$ be the point where $AP$ and $BQ$ meet, and let $R'$ be the point where the line $CX$ meets side $AB$ (or its extension). Since $AP,BQ,CR'$ meet at $X$ (by definition), the theorem in the direction already proved gives $\dfrac{AR'}{R'B}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$. Comparing this with the hypothesis, we get $\dfrac{AR'}{R'B}=\dfrac{AR}{RB}$; but there is only one point that divides side $AB$ in a given ratio, so $R'=R$. That is, $CR$ also passes through the point $X$, and the three lines meet at one point.
Example 2.8: Using Ceva's theorem
On side $AB$ of $\triangle ABC$ take a point $R$ with $AR:RB=2:1$, and on side $BC$ take a point $P$ with $BP:PC=3:1$. If a point $Q$ is taken on side $CA$ so that $AP,BQ,CR$ meet at one point, find $CQ:QA$.
Solution. By Theorem 2.7,
$$ \frac{AR}{RB}\times\frac{BP}{PC}\times\frac{CQ}{QA}=1 $$and substituting $\dfrac{AR}{RB}=\dfrac{2}{1}$ and $\dfrac{BP}{PC}=\dfrac{3}{1}$ into this gives
$$ \frac{2}{1}\times\frac{3}{1}\times\frac{CQ}{QA}=1 $$so $\dfrac{CQ}{QA}=\dfrac{1}{6}$, that is, $CQ:QA=1:6$. (Figure 2.11 was drawn with these values; if we place coordinates such as $A=(2,5)$, $B=(0,0)$, $C=(6,0)$ and actually compute $R,P,Q$, we can verify that $AP,BQ,CR$ meet at exactly the same point $X$.)
2.4.2 Menelaus's theorem
Theorem 2.8: Menelaus's theorem
Suppose that the lines containing the sides $AB,BC,CA$ of $\triangle ABC$ (the sides themselves, or their extensions) meet a line $\ell$ that does not pass through any vertex of the triangle at points $R,P,Q$, respectively. Then
$$ \frac{AR}{RB}\times\frac{BP}{PC}\times\frac{CQ}{QA}=1 $$holds.
Although this looks just like Ceva's theorem, there is a difference: Ceva's theorem is about “three lines passing through the inside of a triangle,” whereas Menelaus's theorem is about “a single line crossing outside a triangle.” The proof also uses a different idea. Incidentally, the mnemonic for this formula can be exactly the same as in the case of Ceva's theorem described in Section 2.4.1 (start from $A$ and go once around: numerator $\to$ denominator, numerator $\to$ denominator, …)—we will check why in Exercise 2.4.
Proof (Theorem 2.8)
We use the following lemma.
Lemma. Suppose that the line through two points $X,Y$ meets a line $\ell$ at a point $Z$. Let $d(X,\ell),d(Y,\ell)$ be the lengths of the perpendiculars dropped from $X,Y$ to $\ell$ (the distances to the feet of the perpendiculars), respectively. Then
$$ \frac{XZ}{ZY}=\frac{d(X,\ell)}{d(Y,\ell)} $$holds.
Proof of the lemma. Let $X',Y'$ be the feet of the perpendiculars dropped from $X,Y$ to $\ell$, respectively. The points $X',Z,Y'$ all lie on the line $\ell$, so they are collinear. $\triangle XZX'$ and $\triangle YZY'$ are right triangles with right angles at $X'$ and $Y'$, respectively, and $\angle XZX'$ and $\angle YZY'$ are vertical angles (angles facing each other across $Z$), so they are equal. Since two pairs of angles are equal, $\triangle XZX'\backsim\triangle YZY'$, and from the ratio of corresponding sides, $\dfrac{XZ}{YZ}=\dfrac{XX'}{YY'}=\dfrac{d(X,\ell)}{d(Y,\ell)}$. ∎
Applying this lemma to the line $\ell$ (the transversal line in the theorem), with the pairs of vertices $(B,C),(C,A),(A,B)$ in turn as $X,Y$, we obtain
$$ \frac{BP}{PC}=\frac{d(B,\ell)}{d(C,\ell)}, \qquad \frac{CQ}{QA}=\frac{d(C,\ell)}{d(A,\ell)}, \qquad \frac{AR}{RB}=\frac{d(A,\ell)}{d(B,\ell)} $$(Here $P,Q,R$ are the points where the lines $BC,CA,AB$, respectively, meet $\ell$.) Multiplying together all three right-hand sides, we get
$$ \frac{BP}{PC}\times\frac{CQ}{QA}\times\frac{AR}{RB}=\frac{d(B,\ell)}{d(C,\ell)}\times\frac{d(C,\ell)}{d(A,\ell)}\times\frac{d(A,\ell)}{d(B,\ell)}=1 $$and the theorem is proved. Note that this lemma holds as it stands whether $Z$ lies inside or outside segment $XY$ (since distances are always taken as positive quantities), so the proof does not change regardless of which of $P,Q,R$ are internal division points of the sides and which are external division points.
∎
Math Note: Using signed ratios gives $-1$
In this book we have treated all of $AR,RB$, and so on as positive lengths and stated the conclusion of the theorem in the form $+1$. More advanced textbooks often use “signed ratios” (taking the value to be negative for an external division) in order to distinguish direction (internal or external division); in that convention the conclusion of the theorem takes the beautiful form $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=-1$, with a minus sign. If you are interested, try checking where this sign comes from using concrete coordinates.
Example 2.9: Using Menelaus's theorem
On side $AB$ of $\triangle ABC$ take a point $R$ with $AR:RB=1:2$, and on side $CA$ take a point $Q$ with $CQ:QA=1:3$. Let $P$ be the point where line $RQ$ meets the extension of side $BC$. Find $BP:PC$.
Solution. By Theorem 2.8,
$$ \frac{AR}{RB}\times\frac{BP}{PC}\times\frac{CQ}{QA}=1 $$Substituting $\dfrac{AR}{RB}=\dfrac{1}{2}$ and $\dfrac{CQ}{QA}=\dfrac{1}{3}$ into this gives
$$ \frac{1}{2}\times\frac{BP}{PC}\times\frac{1}{3}=1 $$so $\dfrac{BP}{PC}=6$; that is, $BP:PC=6:1$. (Figure 2.12 was drawn with these values. If we actually compute the intersection of line $RQ$ and line $BC$ using the coordinates $A=(2,5)$, $B=(0,0)$, $C=(6,0)$, we find that $P$ lies on the extension of $BC$—on the side beyond $C$—and that $BP:PC=6:1$.)
2.5 Summary and Exercises
2.5.1 Summary
- The point that divides segment $BC$ internally in the ratio $m:n$ lies on the segment; the point that divides it externally (with $m\neq n$) lies on the extension of the segment, outside the segment.
- The point where the bisector of $\angle A$ meets side $BC$ divides $BC$ internally in the ratio $AB:AC$ (Theorem 2.1). When $AB\neq AC$, the point where the exterior angle bisector of $\angle A$ meets the extension of side $BC$ divides $BC$ externally in the ratio $AB:AC$ (Theorem 2.2). Both are proofs of the form “draw one auxiliary parallel line, and replace the equal angles made by the bisector with the equal angles made by the parallel line.”
- The centroid (the intersection of the three medians) divides each median internally in the ratio $2:1$, measured from the vertex (Theorem 2.3; it uses the midpoint theorem).
- The median theorem: $AB^2+AC^2=2(AM^2+BM^2)$ ($M$ is the midpoint of $BC$; Theorem 2.4; it uses the Pythagorean theorem).
- The orthocenter (the intersection of the perpendiculars from each vertex to the opposite side) coincides with the circumcenter of the larger triangle that has the original triangle as its medial triangle; from this, its existence (the concurrency of the three perpendiculars) can be proved.
- The circumcenter (the intersection of the perpendicular bisectors of the sides; the center of the circumcircle) and the incenter (the intersection of the bisectors of the angles; the center of the incircle) can both be proved to exist from the idea of “the locus of equidistant points” (Theorem 2.5, Theorem 2.6). The inradius is $r=S/s$ ($S$ is the area and $s$ is the semiperimeter).
- Ceva's theorem: if the three lines $AP,BQ,CR$ meet at a single point, then $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$ (the converse also holds). Menelaus's theorem: the same equation holds for a single line crossing the three sides (or their extensions) of a triangle. The proofs of both are given in this chapter (the original notes omitted the proofs).
2.5.2 Exercises
Exercise 2.1: Interior and exterior angle bisectors
In $\triangle ABC$, suppose $AB=5$, $AC=8$, and $BC=7$. Let $P$ be the point where the bisector of $\angle A$ meets side $BC$, and let $Q$ be the point where the exterior angle bisector of $\angle A$ meets the extension of side $BC$. Find $BP,PC$ and $BQ,CQ$ respectively.
Hint: Use Theorem 2.1 for the internal division and Theorem 2.2 for the external division. Since $AB\lt AC$, be careful about which part of the extension of side $BC$—the $B$ side or the $C$ side—the external division point $Q$ falls on (see the Math Note “How do the positions of $E,Q$ change when $AB\lt AC$?” after the proof of Theorem 2.2).
Exercise 2.2: The five centers of an equilateral triangle
Explain why, in an equilateral triangle, the four points—the centroid, orthocenter, circumcenter, and incenter—all coincide.
Hint: Focus on the symmetry of an equilateral triangle (each median is at the same time a perpendicular bisector, an angle bisector, and a perpendicular from the vertex).
Exercise 2.3: Using the median theorem
In $\triangle ABC$, suppose $AB=7$, $AC=5$, and $BC=8$. Use the median theorem to find the length of the median $AM$ joining vertex $A$ to the midpoint $M$ of side $BC$.
Hint: The computation can be carried out by exactly the same steps as in Example 2.4. The answer is not necessarily an integer.
Exercise 2.4: The difference between Ceva's theorem and Menelaus's theorem
Ceva's theorem and Menelaus's theorem are both expressed by the same equation $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$. Explain how the geometric situations asserted by these two theorems differ. Also explain why the mnemonic “start from $A$ and go once around: numerator $\to$ denominator, numerator $\to$ denominator, …” can be used for both theorems alike.
Hint: Compare the text of Section 2.4 with Figure 2.11 and Figure 2.12. Pay attention to how the arrangement of the points $R,P,Q$ (internal or external division) differs.
Exercise 2.5: Medians and Ceva's theorem
Verify that the three medians of a triangle (the segments drawn from each vertex to the midpoint of the opposite side) satisfy the relation $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$ of Ceva's theorem. Then, using this fact together with Theorem 2.7 (the converse of Ceva's theorem also holds; see the Math Note in Section 2.4.1), give an alternative proof of Theorem 2.3 in Section 2.2 (that the three medians meet at a single point).
Hint: For medians, $R,P,Q$ are all midpoints of the sides, so $AR:RB,\,BP:PC,\,CQ:QA$ are all $1:1$.
Exercise 2.6: Circumcircle and incircle of a right triangle
A right triangle has side lengths $5,12,13$ ($\angle C=90^\circ$, $CA=5,CB=12,AB=13$). Find the circumradius $R$ and the inradius $r$.
Hint: The same idea as in Example 2.6 works for the circumradius, and the same idea as in Example 2.7 works for the inradius.
2.5.3 References
- Yasuhide Mochizuki, Mathematics Notes: High School Mathematics (handwritten lecture notes, in Japanese). The source of this chapter.
- Yoshishige Sugiyama (supervising ed.), Taikei Sūgaku 1 and 2: Kika-hen [Systematic Mathematics 1 and 2: Geometry], Suken Shuppan (in Japanese).
- H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, Mathematical Association of America, 1967.