6. [5] Triangle $A B C$ has $\angle A=90^{\circ}$, side $B C=25, A B>A C$, and area 150 . Circle $\omega$ is inscribed in $A B C$, with $M$ its point of tangency on $A C$. Line $B M$ meets $\omega$ a second time at point $L$. Find the length of segment $B L$.
Answer: $45 \sqrt{17} / 17$. Let $D$ be the foot of the altitude from $A$ to side $B C$. The length of $A D$ is $2 \cdot 150 / 25=12$. Triangles $A D C$ and $B D A$ are similar, so $C D \cdot D B=A D^{2}=144 \Rightarrow B D=16$ and $C D=9 \Rightarrow A B=20$ and $A C=15$. Using equal tangents or the formula inradius as area divided by semiperimeter, we can find the radius of $\omega$ to be 5 . Now, let $N$ be the tangency point of $\omega$ on $A B$. By power of a point, we have $B L \cdot B M=B N^{2}$. Since the center of $\omega$ together with $M, A$, and $N$ determines a square, $B N=15$ and $B M=5 \sqrt{17}$, and we have $B L=45 \sqrt{17} / 17$.