$17 \cdot 71$ Let the two legs of the right $\triangle ABC$ have lengths 5 and 12, respectively. We draw two arcs, one centered at $A$ with a radius of 12, and the other centered at $B$ with a radius of 5. These arcs intersect the hypotenuse at $M$ and $N$, respectively. The length of segment $MN$ is
(A) 2.
(B) $\frac{13}{5}$.
(C) 3.
(D) 4.
(E) $\frac{24}{5}$.
[Solution] By the Pythagorean theorem, $AB = \sqrt{5^2 + 12^2} = 13$, then $MN = AM - AN = AM - (AB - BN) = 4$. Therefore, the answer is $(D)$.