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In triangle $ABC,$ let the angle bisectors be $\overline{BY}$ and $\overline{CZ}$.  Given $AB = 12$, $AY = 12$, and $CY = 6$, find $BC$.

 Dec 18, 2023
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By the angle bisector theorem, [\frac{BY}{CZ} = \frac{AB}{AC}.]

 

We have AY+CY=AC=12+6=18, and BY=AY=12, so CZ=18−12=6.

 

Thus, [\frac{BY}{CZ} = \frac{12}{6} = 2,]so ACAB​=2. Then [\frac{AB}{AC+AB} = \frac{2}{3}.]

 

But by the Law of Sines, [\frac{AB}{AC+AB} = \frac{\sin \angle C}{\sin (A+C)},]so sin(A+C)sin∠C​=32​.

 

Then [\frac{\sin \angle C}{\sin (A+C)} \cdot \frac{\sin (A+C)}{\sin A}

= \frac{2}{3} \cdot \frac{1}{\sin A}.]

 

But sin(A+C)sin∠C​=sin(A+C)sin(180∘−(A+C))​=sin(A+C)sin(A−C)​, so

 

[\frac{\sin (A-C)}{\sin A} = \frac{2}{3}.]

 

By the Law of Sines, [\frac{BC}{\sin A} = \frac{AB}{\sin (A-C)},]

 

so BC=AB⋅sin(A−C)sinA​=12⋅sin(A−C)sinA​=12⋅23​=18​.

 Dec 19, 2023

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