解：解法一：连接

P_{1} P_{2}

，

F H

，

∵ $\frac{P P_{2}}{C H} = \frac{a}{2 a} = \frac{1}{2}$ ， $\frac{P P_{1}}{C F} = \frac{b}{2 b} = \frac{1}{2}$ ，

∴ $\frac{P P_{2}}{C H} = \frac{P P_{1}}{C F}$ .

∵ $∠ P_{1} P P_{2} = ∠ C = 90 °$ ，

∴ $△ P P_{1} P_{2} ∽ △ C F H$ .

∴ $\frac{P_{1} P_{2}}{F H} = \frac{P P_{1}}{C F} = \frac{1}{2}$ ， $\frac{S_{△ P P_{1} P_{2}}}{S_{△ C F H}} = {(\frac{P_{1} P_{2}}{F H})}^{2} = \frac{1}{4}$ .

由（2） $△ P_{1} F Q ∽ △ P_{2} H Q$ ，得 $\frac{P_{1} Q}{P_{2} Q} = \frac{F Q}{H Q}$ ，

∴ $\frac{P_{1} Q}{F Q} = \frac{P_{2} Q}{H Q}$ .

∵ $∠ P_{1} Q P_{2} = ∠ F Q H$ ，

∴ $△ P_{1} Q P_{2} ∽ △ F Q H$ .

∴ $\frac{S_{△ P_{1} Q P_{2}}}{S_{△ F Q H}} = {(\frac{P_{1} P_{2}}{F H})}^{2} = \frac{1}{4}$ .

∵ $S_{1} = S_{△ P P_{1} P_{2}} + S_{△ P_{1} P_{2} Q}$ ，

∴ $S_{1} = \frac{1}{4} S_{△ C F H} + \frac{1}{4} S_{△ F Q M} = \frac{1}{4} (S_{△ C F H} + S_{△ F Q M}) = \frac{1}{4} S_{2}$ .

∴ $\frac{S_{1}}{S_{2}} = \frac{1}{4}$ .

解法二：连接 $P_{1} P_{2}$ 、 $F H$ .

∵ $\frac{P P_{2}}{C H} = \frac{a}{2 a} = \frac{1}{2}$ ， $\frac{P P_{1}}{C F} = \frac{b}{2 b} = \frac{1}{2}$ ，

∴ $\frac{P P_{2}}{C H} = \frac{P P_{1}}{C F}$ .

∵ $∠ P_{1} P P_{2} = ∠ C = 90 °$ ，

∴ $△ P P_{1} P_{2} ∽ △ C F H$ .

∴ $\frac{P_{1} P_{2}}{F H} = \frac{P P_{1}}{C F} = \frac{1}{2}$ ， $∠ P P_{1} P_{2} = ∠ C F H$ ， $∠ P P_{2} P_{1} = ∠ C H F$ .

由（2）中 $△ P_{1} F Q ∽ △ P_{2} H Q$ ，得 $\frac{P_{1} Q}{P_{2} Q} = \frac{F Q}{H Q}$ ，

∴ $\frac{P_{1} Q}{F Q} = \frac{P_{2} Q}{H Q}$ .

∵ $∠ P_{1} Q P_{2} = ∠ F Q H$ ，

∴ $△ P_{1} Q P_{2} ∽ △ F Q H$ .

∴ $\frac{P_{1} Q}{F Q} = \frac{P_{2} Q}{Q H} = \frac{P_{1} P_{2}}{F H} = \frac{1}{2}$ ， $∠ P_{2} P_{1} Q = ∠ H F Q$ ， $∠ P_{1} P_{2} Q = ∠ F H Q$ .

∴ $\frac{P_{1} Q}{F Q} = \frac{P_{2} Q}{H Q} = \frac{P P_{1}}{C F} = \frac{P P_{2}}{C H} = \frac{1}{2}$ ， $∠ P P_{1} Q = ∠ C F Q$ ， $∠ P P_{2} Q = ∠ C H Q$ .

又 $∠ P_{1} P P_{2} = ∠ C$ ， $∠ P_{1} Q P_{2} = ∠ F Q H$ ，

∴四边形 $P P_{1} O P_{2} ∽$ 的四边形 $C F Q H$ .

∴ $\frac{S_{1}}{S_{2}} = {(\frac{P P_{1}}{C F})}^{2} = \frac{1}{4}$

全国历年中考数学真题精选汇编：四边形2