Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Angle Bisector Theorem"

(→‎Introduction)
(→‎Introduction)
Line 6: Line 6:
 
<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>
 
<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 ==
+
== Introduction ==
−
===Method 1 ===
+
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>.  Likewise, the converse of this theorem holds as well.
−
Because of the [[ratio]]s and equal [[angle]]s in the theorem, we think of [[similarity | 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:
 
  
−
<asy>
+
<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>
−
size(200);
 
−
defaultpen(fontsize(10));
 
−
real a,b,c,d,m,n;
 
−
pair A=(1,4), B=(-5,0), C=(3,0), D, E;
 
−
b = abs(C-A);c = abs(B-A);
 
−
D = (b*B+c*C)/(b+c);
 
−
m = abs(B-D);n = abs(C-D);
 
−
E = C+(B-A)*n/m;
 
−
draw(A--B--C--A--E--C);
 
−
MA(B,A,D,0.5,blue+linewidth(1));
 
−
MA(D,A,C,0.6,blue+linewidth(1));
 
−
MA(C,E,A,0.6,blue+linewidth(1));
 
−
MA(C,B,A,0.6,green+linewidth(1));
 
−
MA(B,C,E,0.6,green+linewidth(1));
 
−
label("$A$",A,(1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(1,-1));label("$E$",E,(0,-1));
 
−
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));label("$b$",(E+C)/2,(1,-0.5));
 
−
dot(A^^B^^C^^D^^E);
 
−
</asy>
 
−
 
 
−
Since AB and CE are parallel, we know that <math> \angle BAE=\angle CEA </math> and <math> \angle BCE=\angle ABC </math>.  Triangle ACE is [[isosceles triangle | isosceles]], with AC = CE.
 
−
 
 
−
By [[AA similarity]], <math> \triangle DAB \cong \triangle DEC </math>.  By the properties of similar triangles, we arrive at our desired result:
 
−
 
 
−
<center><math> \frac cm = \frac bn.</math> </center>
 
−
 
 
−
=== Method 2 ===
 
−
Since B,D, and C are collinear, <math>\dfrac{[ABD]}{[ADC]} = \dfrac{m}{n}</math>. Now consider the perpendicular from <math>D</math> to <math>AB</math> and <math>D</math> to <math>AC</math>. Every point on the angle bisector of an angle is equidistant to the sides of the angle, so the height <math>h</math> to <math>AB</math> is equal to the height to <math>AC</math>. Thus <math>\dfrac{[ABD]}{[ADC]} = \dfrac{\dfrac{ch}{2}}{\dfrac{bh}{2}} = \dfrac{c}{b}</math>. Thus <math>\dfrac{c}{b} = \dfrac{m}{n}</math>, so <math>\dfrac{c}{m} = \dfrac{b}{n}</math>. We can prove the converse by the Phantom Point Method, since we can find <math>m</math> and <math>n</math> in terms of <math>a</math>, <math>b</math>, and <math>c</math>, and prove that the points are the same.
 
−
 
 
−
=== Method 3 ===
 
−
Let <math>AD = d</math>. Now, we can express the area of triangle ABD in two ways:
 
−
 
 
−
<math>[ABD] = \frac 12 cd\sin \angle BAD = \frac 12  md \sin \angle ADB.</math>
 
−
Thus, <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm</math>.
 
−
 
 
−
Likewise, triangle ACD can be expressed in two different ways:
 
−
 
 
−
<math>[ACD] = \frac 12 bd \sin \angle CAD = \frac 12 dn \sin \angle ADC.</math>
 
−
Thus, <math>\frac{\sin \angle ADC}{\sin \angle CAD} = \frac bn</math>.
 
−
 
 
−
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>.
 
−
 
 
−
We conclude that <math>\frac{\sin \angle ADB}{\sin \angle BAD} = \frac cm = \frac bn</math>, which was what we wanted.
 
−
 
 
−
In both cases, if we reverse all the steps, we see that everything still holds and thus the converse holds.
 
  
 
== Examples ==
 
== Examples ==

Revision as of 22:56, 17 August 2016

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

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

[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]

Introduction

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

[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]

Examples

  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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&oldid=80010"