Difference between revisions of "Angle Bisector Theorem"
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Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that | Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that | ||
sin(BDA)=sin(ADC) | sin(BDA)=sin(ADC) | ||
− | 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. | + | 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. ~ sachi2019 |
== Examples == | == Examples == |
Revision as of 13:49, 24 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 LoS on ACD and ABD, AB/BD=sin(BDA)/sin(BAD) ... (1) AC/AD=sin(ADC)/sin(DAC) ... (2) Well, we also know that BDA and ADC add to 180 degrees. I think that means that we can use sin(180-x)=sin(x) here. Doing so, we see that sin(BDA)=sin(ADC) 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. ~ sachi2019
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.