$17 \cdot 76$ Given a right triangle $ABC$, with the two legs $BC=3, AC=4$, find the length of the shorter angle bisector that trisects the right angle.
(A) $\frac{32 \sqrt{3}-24}{13}$.
(B) $\frac{12 \sqrt{3}}{13}-9$.
(C) $6 \sqrt{3}-8$.
(D) $\frac{5 \sqrt{10}}{6}$.
(E) $\frac{25}{12}$.
(11th American High School Mathematics Examination, 1960)
[Solution] Let $CD$ be the trisector of $\angle C$, and draw $DE \perp AC$ at $E$. Let $EC = x$.

In $\mathrm{Rt} \triangle DEC$, $\angle CDE = 30^{\circ}$, so $CD = 2x$, $DE = \sqrt{3} x$.
By the property of similar triangles, we have $\quad \frac{AE}{DE} = \frac{AC}{BC}$,
which is $\frac{4-x}{\sqrt{3} x} = \frac{4}{3}$,
solving for $x$ gives $x = \frac{12}{3+4 \sqrt{3}} = \frac{16 \sqrt{3}-12}{13}$.
Therefore, the answer is $(A)$.