Difference between revisions of "Angle bisector theorem"
Etmetalakret (talk | contribs) (Removed redirect to Angle Bisector Theorem) (Tag: Removed redirect) |
(→Proof) |
||
Line 16: | Line 16: | ||
\frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}</cmath> | \frac{AC}{CD}&=\frac{\sin(ADC)}{\sin(CAD)}\end{align*}</cmath> | ||
− | First, because <math>\ | + | 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. |
Second, we observe that <math>m\angle BDA + m\angle CDA = \pi</math> and <math>\sin(\pi - \theta) = \sin(\theta)</math>. | Second, we observe that <math>m\angle BDA + m\angle CDA = \pi</math> and <math>\sin(\pi - \theta) = \sin(\theta)</math>. |
Latest revision as of 14:20, 4 March 2023
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 the Law of Sines on and
,
First, because is an angle bisector, we know that
and thus
, so the denominators are equal.
Second, we observe that and
.
Therefore,
, so the numerators are equal.
It then follows that
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.