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Difference between revisions of "Angle Bisector Theorem"
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== Introduction ==
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{{WotWAnnounce|week=June 6-12}}
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The '''Angle Bisector Theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC,  then <math> \frac cm = \frac bn </math>.
 
  
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<center>[[Image:Anglebisector.png]]</center>
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== Introduction & Formulas ==
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The '''Angle bisector theorem''' states that given [[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC,  then <math> \frac cm = \frac bn </math>. It follows that <math> \frac cb = \frac mn </math>. Likewise, the [[converse]] of this theorem holds as well.
  
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== Proof ==
 
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===Method 1 ===
 
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Because of the ratios and equal angles in the theorem, we think of similar triangles.  There are not any similar triangles in the figure as it now stands, however.  So we think to draw in a carefully chosen line or two.  Extending AD until it hits the line through C parallel to AB does just the trick as we shall see!
 
  
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<center>[[image:Anglebisectortheorem.PNG]]</center>
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Further by combining with [[Stewart's theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math>
  
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Since AB and CE are parallel, we know that <math> \angle BAE \cong \angle CEA </math> and <math> \angle BCE \cong \angle ABC </math>.  Triangle ACE is isosceles meaning that AC = CE.
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<asy> size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy>
  
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By AA, <math> \triangle DAB \cong \triangle DEC </math>.  By the properties of similar triangles, we arrive at our desired result:
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== Proof ==
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<center><math> \frac cm = \frac bn.</math> </center>
 
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=== Method 2 ===
 
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Let <math> AD = d </math>.  Now, we can express the area of triangle ABD in two ways:
 
  
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<center><math> [ABD] = \frac 12 cd\sin \angle BAD = \frac 12  md \sin \angle ADB. </math></center>
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By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</math>,
  
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Thus <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm </math>.
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<cmath>\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\
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\frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}</cmath>
  
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Likewise, triangle ACD can be expressed in two different ways:
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First, because <math>\overline{AD}</math> is an angle bisector, we know that <math>m\angle BAD = m\angle CAD</math> and thus <math>\sin(BAD) = \sin(CAD)</math>, so the denominators are equal.
  
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<center><math> [ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC. </math></center>
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Second, we observe that <math>m\angle BDA + m\angle CDA = \pi</math> and <math>\sin(\pi - \theta) = \sin(\theta)</math>.
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Therefore, <math>\sin(BDA) = \sin(CDA)</math>, so the numerators are equal.
  
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Thus <math> \frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn</math>.
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It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}</cmath>
  
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But <math> \angle CAD \cong \angle BAD </math> and <math> \sin \angle ADC = \sin \angle ADB </math> since <math> \angle ADC = \pi - \angle ADB </math>.  Therefore, we can substitute back into our previous equation to get <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac bn </math>.
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== Examples & Problems ==
  
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We conclude that <math> \frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn </math>, which was what we wanted.
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# Let ABC be a triangle with angle bisector AD with D on line segment BC.  If <math> BD = 2, CD = 5,</math> and <math> AB + AC = 10 </math>, find AB and AC.<br> '''''Solution:''''' By the angle bisector theorem, <math> \frac{AB}2 = \frac{AC}5</math> or <math> AB = \frac 25 AC </math>.  Plugging this into <math> AB + AC = 10 </math> and solving for AC gives <math> AC = \frac{50}7</math>.  We can plug this back in to find <math> AB = \frac{20}7 </math>.
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# In triangle ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>.  Find the value of <math> m\angle BAP - m\angle CAP </math>. <br> '''''Solution:'''''  First, we notice that <math> \frac{AB}{BP}=\frac{AC}{CP} </math>.  Thus, AP is the angle bisector of angle A, making our answer 0.
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== Examples ==
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# Part '''(b)''', [[1959 IMO Problems/Problem 5]]
  
 
== See also ==
 
== See also ==
 
* [[Angle bisector]]
 
* [[Angle bisector]]
 
* [[Geometry]]
 
* [[Geometry]]
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* [[Stewart's Theorem]]
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* [[Stewart's theorem]]
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[[Category:Geometry]]
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[[Category:Theorems]]

Latest revision as of 00:58, 20 May 2025

This is an AoPSWiki Word of the Week for June 6-12

Introduction & Formulas

The Angle bisector theorem states that given triangle $\triangle ABC$ and angle bisector AD, where D is on side BC, then $\frac cm = \frac bn$. It follows that $\frac cb = \frac mn$. Likewise, the converse of this theorem holds as well.


Further by combining with Stewart's theorem it can be shown that $AD^2 = b\cdot c - m \cdot n$

[asy] size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1)); dot(A^^B^^C^^D,blue);label("$b$",(A+C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); [/asy]

Proof

By the Law of Sines on $\angle ACD$ and $\angle ABD$,

\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\ \frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}

First, because $\overline{AD}$ is an angle bisector, we know that $m\angle BAD = m\angle CAD$ and thus $\sin(BAD) = \sin(CAD)$, so the denominators are equal.

Second, we observe that $m\angle BDA + m\angle CDA = \pi$ and $\sin(\pi - \theta) = \sin(\theta)$. Therefore, $\sin(BDA) = \sin(CDA)$, so the numerators are equal.

It then follows that \[\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}\]

Examples & Problems

  1. Let ABC be a triangle with angle bisector AD with D on line segment BC. If $BD = 2, CD = 5,$ and $AB + AC = 10$, find AB and AC.
    Solution: By the angle bisector theorem, $\frac{AB}2 = \frac{AC}5$ or $AB = \frac 25 AC$. Plugging this into $AB + AC = 10$ and solving for AC gives $AC = \frac{50}7$. We can plug this back in to find $AB = \frac{20}7$.
  2. In triangle ABC, let P be a point on BC and let $AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3$. Find the value of $m\angle BAP - m\angle CAP$.
    Solution: First, we notice that $\frac{AB}{BP}=\frac{AC}{CP}$. Thus, AP is the angle bisector of angle A, making our answer 0.
  3. Part (b), 1959 IMO Problems/Problem 5

See also

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