<s>[INST] △ with altitude from  and angle bisector of  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.
 [/INST] 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, $a+b+c= 47$.
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