25. [13] In convex quadrilateral $A B C D$ with $A B=11$ and $C D=13$, there is a point $P$ for which $\triangle A D P$ and $\triangle B C P$ are congruent equilateral triangles. Compute the side length of these triangles.
Answer:
7
Solution:

Evidently $A B C D$ is an isosceles trapezoid with $P$ as its circumcenter. Now, construct isosceles trapezoid $A B^{\prime} B C$ (that is, $B B^{\prime}$ is parallel to $A C$.) Then $A B^{\prime} P D$ is a rhombus, so $\angle B^{\prime} C D=\frac{1}{2} \angle B^{\prime} P D=60^{\circ}$ by the inscribed angle theorem. Also, $B^{\prime} C=11$ because the quadrilateral $B^{\prime} A P C$ is a $60^{\circ}$ rotation of $A D P B$ about $P$. Since $C D=13$, we use the law of cosines to get that $B^{\prime} D=7 \sqrt{3}$. Hence $A P=7$.