17・120 $P$ is any point on side $BC$ of equilateral $\triangle ABC$, $PX \perp AB, PY \perp AC$, connect $XY$, and let the perimeter of $\triangle XAY$ be $L$, and the perimeter of quadrilateral $BCYX$ be $S$, then the relationship between $L$ and $S$ is
(A) $S>L$.
(B) $S=L$.
(C) $S<L$.
(D) Uncertain.
(Chinese Zhejiang Province Junior High School Mathematics Competition, 1992)
[Solution] Let the side length of equilateral $\triangle ABC$ be $a$. Given $BX = \frac{1}{2} BP$ and $YC = \frac{1}{2} PC$ (both $\triangle P X B$ and $\triangle P Y C$ are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles).
Then $BX + BC + CY = BC + \frac{1}{2}(BP + PC) = \frac{3}{2} a$.
And $AX + AY = 2a - (BX + CY) = 2a - \frac{a}{2} = \frac{3}{2} a$.
$\therefore AX + AY + XY = BX + BC + CY + XY$, i.e., $S = L$. Therefore, the answer is $(B)$.