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Difference between revisions of "Angle Bisector Theorem"

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#REDIRECT[[Angle bisector theorem]]
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{{WotWAnnounce|week=June 6-12}}
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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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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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<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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== Proof ==
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By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</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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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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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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It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{CD}</cmath>
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== Examples & Problems ==
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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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# Part '''(b)''', [[1959 IMO Problems/Problem 5]].
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== See also ==
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* [[Angle bisector]]
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* [[Geometry]]
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* [[Stewart's theorem]]
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[[Category:Geometry]]
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[[Category:Theorems]]

Latest revision as of 12:17, 30 November 2024

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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