Difference between revisions of "Angle Bisector Theorem"
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== Proof == | == Proof == | ||
− | By LoS on | + | By <math>LoS</math> on <math>\angleACD</math> and <math>\angleABD</math>, |
− | AB | + | <math>\frac{AB}{BD}=\frac{sin(BDA)}{sin(BAD)}</math> ... <math>(1)</math> and |
− | AC | + | <math>\frac{AC}{AD}=\frac{sin(ADC)/sin(DAC)} </math>... <math>(2)</math> |
− | Well, we also know that BDA and ADC add to 180 | + | 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> |
− | sin(BDA)=sin(ADC) | + | 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. ~ SilverLightning59 |
− | I noticed that these are the numerators of (1) and (2) respectively. Since BAD and DAC are equal, then you get the equation for the bisector angle theorem. ~ | ||
== Examples == | == Examples == |
Revision as of 04:17, 26 April 2020
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Contents
Introduction
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 $\angleACD$ (Error compiling LaTeX. ! Undefined control sequence.) and $\angleABD$ (Error compiling LaTeX. ! Undefined control sequence.),
... and $\frac{AC}{AD}=\frac{sin(ADC)/sin(DAC)}$ (Error compiling LaTeX. ! Missing } inserted.)... 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
- 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.