$17 \cdot 70$ In a right triangle, given $B M+M A=B C+C A$, if $M B=x$, $C B=h, C A=d$, then $x$ equals
(A) $\frac{h d}{2 h+d}$.
(B) $d-h$.
(C) $\frac{1}{2} d$.
(D) $h+d=\sqrt{2} d$.
(E) $\sqrt{h^{2}+d^{2}}=h$.
(5th American High School Mathematics Examination, 1954)
［Solution］By the given and Pythagorean theorem, we have
$$
M A^{2}=(x+h)^{2}+d^{2},
$$

Also, $M A+x=h+d$, so $M A=h+d-x$,
$\therefore \quad(h+d-x)^{2}=(x+h)^{2}+d^{2}$,
we get
$$
x=\frac{h d}{2 h+d} \text {. }
$$

Therefore, the answer is $(A)$.