Difference between revisions of "Angle Bisector Theorem"
(→Introduction) |
(→Proof) |
||
Line 16: | Line 16: | ||
<math>\frac{AC}{AD}=\frac{sin(ADC)}{sin(DAC)}</math>... <math>(2)</math> | <math>\frac{AC}{AD}=\frac{sin(ADC)}{sin(DAC)}</math>... <math>(2)</math> | ||
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> | 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. | + | 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. |
== Examples & Problems == | == Examples & Problems == |
Revision as of 04:24, 26 April 2020
This is an AoPSWiki Word of the Week for June 6-12 |
Introduction & Formulas
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then
. It follows that
. Likewise, the converse of this theorem holds as well.
Further by combining with Stewart's Theorem it can be shown that
Proof
By on
and
,
...
and
...
Well, we also know that
and
add to
. I think that means that we can use
here. Doing so, we see that
I noticed that these are the numerators of
and
respectively. Since
and
are equal, then you get the equation for the bisector angle theorem.
Examples & Problems
- Let ABC be a triangle with angle bisector AD with D on line segment BC. If
and
, find AB and AC.
Solution: By the angle bisector theorem,or
. Plugging this into
and solving for AC gives
. We can plug this back in to find
.
- In triangle ABC, let P be a point on BC and let
. Find the value of
.
Solution: First, we notice that. Thus, AP is the angle bisector of angle A, making our answer 0.
- Part (b), 1959 IMO Problems/Problem 5.