# △𝐴𝐵𝐶 with altitude from 𝐴 and angle bisector of 𝐵 intersecting at $P$. Find $BP$

Let $$\triangle ABC$$ be a triangle with side lengths $$AB = 17, BC = 28, AC = 25$$. Let the altitude from $$A$$ to $$BC$$ and the angle bisector of angle $$B$$ meet at $$P$$. Given the length of $$BP$$ can be expressed as $$\frac{a\sqrt{b}}{c}$$ for positive integers $$a, b, c$$ where $$gcd(a, c) = 1$$ and $$b$$ is not divisible by the square of any prime, find $$a + b + c$$.

I drew my diagram out, and assumed that the intersection point is the midpoint of the altitude. Then i used pythag to find the length of the segment where the altitude intersects BC. I got a huge answer which is 99% incorrect so i need help on solving it.

Let $$\angle B=2\beta$$ and $$D$$ be the foot of the altitude from $$A$$. Then $$\cos\beta = \frac{BD}{BP}, \>\>\> \cos 2\beta = \frac{BD}{17}$$ leading to $$BP= \frac{17}{\cos\beta}(2\cos^2\beta-1)\tag1$$ From the cosine rule $$\cos2\beta= \frac{ 17^2+28^2-25^2}{2\cdot 17\cdot 28}=2\cos\beta^2-1$$ which yields $$\cos^2\beta=\frac{25}{34}$$. Plug into (1) to obtain $$BP= \frac{8\sqrt{34}}5$$ Thus, $$a+b+c= 47$$.