延长EP至F交AB于F由BP⊥PQ易证△EPQ∽△FBP由BP⊥PQ,BC⊥CQ则B,C,Q,P四点共圆∠PBQ=∠ACD△BPQ∽△CDA则PQ/BP=DA/CD=1/(2√2)S△EPQ/S△FBP=[1/(2√2)]^2=1/8在△FBP中,PF=x/3,BF=2√2-2√2x/3S△FBP=1/2PF*BFS△EPQ=1/8S△FBP=1/16PF*BF=√2/72x(3-x)
其中0<x≤8/3,8/3是在BP⊥AC时取
由y=√2/72x(3-x)为开口向下的抛物线
当x=3-x时即x=3/2取最大值为√2/32... |
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