16. Let $D$ be a point inside $\triangle A B C$ such that $\angle B A D=\angle B C D$ and $\angle B D C=90^{\circ}$. If $A B=$ $5, B C=6$ and $M$ is the midpoint of $A C$, find the length of $D M$.
(2 marks)
設 $D$ 為 $\triangle A B C$ 內的一點, 使得 $\angle B A D=\angle B C D$ 及 $\angle B D C=90^{\circ}$ 。若 $A B=5$, $B C=6$, 且 $M$ 是 $A C$ 的中點, 求 $D M$ 的長度。
16. $\frac{\sqrt{11}}{2}$
16. Extend $C D$ to $E$ so that $D E=D C$. Then $\triangle B D E$ and $\triangle B D C$ are congruent so that we have $B E=B C=6$ and $\angle B E D=\angle B C D=\angle B A D$. It follows that $A, D, B, E$ are concyclic and so $\angle E A B=\angle E D B=90^{\circ}$. Thus $A E=\sqrt{6^{2}-5^{2}}=\sqrt{11}$ and hence the mid-point theorem asserts that $D M=\frac{1}{2} A E=\frac{\sqrt{11}}{2}$.