11. In the diagram below, the point $D$ lies inside the triangle $A B C$ such that $\angle B A D=\angle B C D$ and $\angle B D C=90^{\circ}$. Given that $A B=5$ and $B C=6$, and the point $M$ is the midpoint of $A C$, find the value of $8 \times D M^{2}$.
11. Answer: 22
Solution. Extend $C D$ to $E$ such that $C D=D E$.

It is clear that $\triangle C D B$ and $\triangle E D B$ are congruent. Hence $E B=C B=6$ and $\angle B E D=$ $\angle B C D$. Thus, $\angle B E D=\angle B C D=\angle B A D$ implies that the points $B, D, A$ are $E$ are concyclic. Given that $\angle B D C=90^{\circ}$, hence $\angle E D B=90^{\circ} . B D A E$ is a cyclic quadrilateral with $E B$ as a diameter. Thus, $\angle E A B=90^{\circ}$. In the right-angled triangle $E A B$, we have
$$
A E=\sqrt{E B^{2}-A B^{2}}=\sqrt{6^{2}-5^{2}}=\sqrt{11} .
$$

Since $D$ and $M$ are the midpoints of $E C$ and $A C$ respectively, $D M=\frac{1}{2} A E=\frac{\sqrt{11}}{2}$. Thus, $8 \times D M^{2}=22$