Difference between revisions of "Angle Bisector Theorem"

Revision as of 22:08, 12 June 2020

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}{AD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}

First, because $\bar{AD}$ 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 $\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{AD}$

Examples & Problems

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 $AB = \frac{20}7$ .
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.
Part (b), 1959 IMO Problems/Problem 5.

@@ Line 11: / Line 11: @@
 == Proof ==
-By <math>LoS</math> on <math>\angle ACD</math> and <math>\angle ABD</math>,
+By the [[Law of Sines]] on <math>\angle ACD</math> and <math>\angle ABD</math>,
-<math>\frac{AB}{BD}=\frac{sin(BDA)}{sin(BAD)}</math> ... <math>(1)</math> and
+\begin{align*}\frac{AB}{BD}&=\frac{\sin(BDA)}{\sin(BAD)}\\
-<math>\frac{AC}{AD}=\frac{sin(ADC)}{sin(DAC)}</math>... <math>(2)</math>
+\frac{AC}{AD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}
-Well, we also know that <math>\angle BDA</math> and <math>\angle ADC</math> add to <math>180^\circ</math>. I think that means that we can use <math>sin(180-x)=sin(x)</math> here. Doing so, we see that <math>sin(BDA)=sin(ADC)</math>
-I noticed that these are the numerators of <math>(1)</math> and <math>(2)</math> respectively. Since <math>\angle BAD</math> and <math>\angle DAC</math> are equal, then you get the equation for the bisector angle theorem.
+First, because <math>\bar{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.
+Second, we observe that <math>m\angle BDA + m\angle CDA = \pi</math> and <math>\sin(\pi - \theta) = \sin(\theta)</math>.
+Therefore, <math>\sin(BDA) = \sin(CDA)</math>, so the numerators are equal.
+It then follows that <cmath>\frac{AB}{BD}=\frac{\sin(BDA)}{\sin(BAD)} = \frac{AC}{AD}</cmath>
 == Examples & Problems ==

Page

Toolbox

Search

Difference between revisions of "Angle Bisector Theorem"

Revision as of 22:08, 12 June 2020

Contents

Introduction & Formulas

Proof

Examples & Problems

See also