Difference between revisions of "Angle Bisector Theorem"
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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. | 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. | ||
Revision as of 04:23, 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. ~ SilverLightning59
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.