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# BY and CZ are angle bisectors of triangle $ABC$ that meet at $I$ as shown below, with $CY = 4$, $AY = 6$, and $AB = 8$. Find $BC$. ​

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BY and CZ are angle bisectors of triangle $ABC$ that meet at $I$ as shown below, with $CY = 4$, $AY = 6$, and $AB = 8$. Find $BC$.

Apr 14, 2020

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To solve this problem, I will first introduce a interesting theorem.

Theorem ABT:

If Y is a point on AC such that BY bisects $$\angle ABC$$, then $$\dfrac{AY}{YC} = \dfrac{AB}{BC}$$.

Proof of Theorem ABT:

Let $$\angle CYB = \alpha$$$$\angle ABY = \beta$$.

Then $$\angle AYB = 180^\circ - \alpha$$

$$\sin \angle CYB = \sin \angle AYB = \sin \alpha$$

Applying Law of Sines on $$\triangle AYB$$,

$$\dfrac{AY}{AB} = \dfrac{\sin \beta}{\sin \alpha}$$

Applying Law of Sines again on $$\triangle CYB$$,

$$\dfrac{YC}{BC} = \dfrac{\sin \beta}{\sin \alpha}$$

Therefore,

$$\dfrac{AY}{AB} = \dfrac{YC}{BC}\\ \dfrac{AY}{YC} = \dfrac{AB}{BC}$$

Back to the problem. We apply Theorem ABT.

$$\dfrac{AY}{YC} = \dfrac{AB}{BC}\\ \dfrac{6}{4} = \dfrac{8}{BC}\\ BC = \dfrac{16}{3}$$

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Apr 14, 2020