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 $BC$. ​

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.

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