?$證明： (1)因?yàn)镻E//DQ$?
?$所以∠APE=∠ADQ，∠AEP=∠AQD$?
?$所以△APE∽△ADQ$?
?$(2)由(1)△APE∽△ADQ$?
?$同理可得△PDF∽△ADQ$?
?$相似比分別為\frac {x}{3}，\frac {3-x}{3}$?
?$所以面積比分別為\frac {x2}{9}.\frac {(3-x)2}{9}$?
?$S_{△ADQ} =\frac {1}{2}×2×3=3$?
?$所以S_{△APE}=\frac {x2}{3}，S_{△PDF}=\frac {(3-x)2}{3}$?
?$所以S_{平行四邊形PEQF}= 2S= 3-\frac {x2}{3}-\frac {(3-x)2}{3}$?
?$所以S= -\frac {1}{3}x2+x$?
?$S= -\frac {1}{3}(x-\frac {3}{2})2+\frac {3}{4}$?
?$當(dāng)x=\frac {3}{2}即AP的長(zhǎng)為\frac {3}{2}時(shí)， S取得最大值，最大值是\frac {3}{4}$
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