25. A pentagon $A B C D E$ is inscribed in a circle of radius 10 such that $B C$ is parallel to $A D$ and $A D$ intersects $C E$ at $M$. The tangents to this circle at $B$ and $E$ meet the extension of $D A$ at a common point $P$. Suppose $P B=P E=24$ and $\angle B P D=30^{\circ}$. Find $B M$.
25. Ans: 13
Let $O$ be the centre of the circle. First we show that $P, B, O, M, E$ lie on a common circle. Clearly $P, B, O, E$ are concyclic. As $\angle B P M=\angle H B C=\angle B E M=30^{\circ}$, points $P, B, M, E$ are also concyclic. Thus $P, B, O, M, E$ all lie on the circumcircle of $\triangle P B E$.

Since $\angle P B O=90^{\circ}, P O$ is a diameter of this circle and $P O=\sqrt{24^{2}+10^{2}}=26$.
This circle is also the circumcircle of $\triangle P B M$. Therefore, $B M=26 \sin 30^{\circ}=13$.