1,若ΔABC与ΔDAP相似，则∠APD=∠BCA=∠C=60o;2,设PC=X，PD‖BA,∠PDC=∠BAC=90o=∠PDA，∠CPD=∠B=30o，AC=BC/2=24/2=12,DC=PC/2=X/2,PD=PC√3/2=X√3/2,SΔAPD=AD*PD/2=(AC-DC)*PD/2=(12-X/2)(X√3/2)/2=3X√3-X2√3/8=-(√3/8)(X2-24X)=-(√3/8)[(X-12)2-122]=18√3-(√3/8)(X-12)2(X-12)2>=0,-(√3...
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