Question

如图，P,Q是矩形ABCD的边BC和CD延长线上的点，连结AP,AQ,PQ,AP与CQ相交于点E，且∠PAD=∠QAD,求证S矩形ABC

Answer 1

方法1：延长AD交PQ于F,S三角形APQ=1/2*AF*DQ+1/2*AF*CD=1/2*AF*CQ 设AB=a,BC=b,DQ=DE=h 则AD:CP=DE:CE 即b/CP=h/(a-h) CP=b(a-h)/h 且DF:CP=DQ:CQ 即DF/CP=h/(h+a) 推得:DF=b*(a-h)/(a+h) 所以AF=b+DF=2ab/(a+h) S三角形APQ=1/2*2ab/(a+h)*(a+h)=ab=S矩形ABCD

方法2：因为∠QAP=∠PAD 可得D为QE中点 S△AQP=S△QAE+S△QEP S△QEP=CP*QE/2 CP/CE=AD/DE 所以S△QEP=AD*CE*QE/2DE=AD*CE S△APQ=2S△ADE+S△PEQ=2AD*DE/2+AD*CE=AD*(DE+CE)=S矩形ABCD