Triangle $\triangle ABC$ has side lengths $AB = 80$, $BC = 45$, and $AC = 75$. The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P.$ What is $BP$? $\textbf{(A)}~18\qquad$ $\textbf{(B)}~19\qquad$ $\textbf{(C)}~20\qquad$ $\textbf{(D)}~21\qquad$ $\textbf{(E)}~22$
Solution
By the law of cosines, $\cos B=\dfrac{80^2+45^2-75^2}{2\cdot80\cdot45}=\dfrac{2800}{7200}=\dfrac7{18}$.
Let $H$ be the foot of the altitude from $C$ to $\overline{AB}$. In the right triangle $BHC$, $BH=BC\cos B=45\cdot\dfrac7{18}=\dfrac{35}2$.
The point $P$ lies on $\overline{CH}$ and on the bisector of $\angle B$, so in the right triangle $BHP$ we have $\angle PBH=\frac B2$ and $BP=\dfrac{BH}{\cos(B/2)}$.
$\cos^2\frac B2=\dfrac{1+\cos B}2=\dfrac{25}{36}$, so $\cos\frac B2=\dfrac56$. Hence $BP=\dfrac{35/2}{5/6}=21$.
In $\triangle ABC$, $AB=18$, $BC=30$, and $AC=33$. The bisector of $\angle B$ and the altitude from $C$ to line $AB$ intersect at point $P$. What is $BP$?
