# A triangle has corners at points A, B, and C. Side AB has a length of 48 . The distance between the intersection of point A's angle bisector with side BC and point B is 9 . If side AC has a length of 36 , what is the length of side BC?

Nov 10, 2016

$\frac{63}{4}$

#### Explanation:

Call $P$ the intersection point in the question.
Then using the sin theorem applied on triangles APC and APB we get

$\frac{36}{\sin} \left(A \hat{P} C\right) = \frac{\overline{P C}}{\sin} \left(P \hat{A} C\right)$

$\frac{48}{\sin} \left(\pi - A \hat{P} C\right) = \frac{9}{\sin} \left(P \hat{A} B\right)$

$P \hat{A} C = P \hat{A} B \setminus \setminus \setminus$ by consdtruction and

$\sin \left(A \hat{P} C\right) = \sin \left(\pi - A \hat{P} C\right)$

$\frac{\overline{P C}}{36} = \sin \frac{P \hat{A} C}{\sin} \left(A \hat{P} C\right) = \sin \frac{P \hat{A} B}{\sin} \left(\pi - A \hat{P} C\right) = \frac{9}{48}$

so

$\overline{P C} = 9 \cdot \frac{36}{48} = 9 \cdot \frac{3}{4}$

and

$\overline{B C} = \overline{B P} + \overline{P C} = 9 + \frac{27}{4} = \frac{36 + 27}{4} = \frac{63}{4}$