Let triangle $A B C$ be such that $A B=A C=22$ and $B C=11$. Point $D$ is chosen in the interior of the triangle such that $A D=19$ and $\angle A B D+\angle A C D=90^{\circ}$. The value of $B D^{2}+C D^{2}$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$.
Solution: Rotate triangle $A B D$ about $A$ so that $B$ coincides with $C$. Let $D$ map to $D^{\prime}$ under this. Note that $C D D^{\prime}$ is a right triangle with right angle at $C$. Also, note that $A D D^{\prime}$ is similar to $A B C$. Thus, we have $D D^{\prime}=\frac{A D}{2}=\frac{19}{2}$. Finally, note that \[ B D^{2}+C D^{2}=C D^{\prime 2}+C D^{2}=D D^{\prime 2}=\frac{361}{4} \] $\fbox{36104}$.