$17 \cdot 178$ On the hypotenuse $AB$ of the isosceles right triangle $ABC$, take two points $M$ and $N$ such that $\angle MCN=45^{\circ}$. Let $AM=m, MN=x, BN=n$. Then the shape of the triangle with side lengths $x, m, n$ is
(A) Acute triangle.
(B) Right triangle.
(C) Obtuse triangle.
(D) Varies with $x, m, n$.
(Anhui Province, China Junior High School Mathematics Competition, 1997)
[Solution] $\because \angle M C N=\angle A=\angle B=45^{\circ}$,
$\therefore \triangle C M N \backsim \triangle A C N \backsim \triangle B M C$.
$\therefore \quad \frac{B C}{x+n}=\frac{m+x}{A C}$.
$\because \quad B C=A C . \quad \therefore \quad B C^{2}=(x+n)(x+m)$.
Also, $2 B C^{2}=A B^{2}, \therefore 2(x+m)(x+n)=(m+x+n)^{2}$,
Simplifying, we get
$x^{2}=m^{2}+n^{2}$.

Therefore, the answer is $(B)$.