8. [7] Triangle $A B C$ has side lengths $A B=231, B C=160$, and $A C=281$. Point $D$ is constructed on the opposite side of line $A C$ as point $B$ such that $A D=178$ and $C D=153$. Compute the distance from $B$ to the midpoint of segment $A D$.
Answer:
208
Solution: Note that $\angle A B C$ is right since
$$
B C^{2}=160^{2}=50 \cdot 512=(A C-A B) \cdot(A C+A B)=A C^{2}-A B^{2} .
$$

Construct point $B^{\prime}$ such that $A B C B^{\prime}$ is a rectangle, and construct $D^{\prime}$ on segment $B^{\prime} C$ such that $A D=A D^{\prime}$. Then
$$
B^{\prime} D^{\prime 2}=A D^{\prime 2}-A B^{\prime 2}=A D^{2}-B C^{2}=(A D-B C)(A D+B C)=18 \cdot 338=78^{2} .
$$

It follows that $C D^{\prime}=B^{\prime} C-B^{\prime} D^{\prime}=153=C D$; thus, points $D$ and $D^{\prime}$ coincide, and $A B \| C D$. Let $M$ denote the midpoint of segment $A D$, and denote the orthogonal projections $M$ to lines $A B$ and $B C$ by $P$ and $Q$ respectively. Then $Q$ is the midpoint of $B C$ and $A P=39$, so that $P B=A B-A P=192$ and
$$
B M=P Q=\sqrt{80^{2}+192^{2}}=16 \sqrt{5^{2}+12^{2}}=208 .
$$