如图,梯形ABCD中,AD∥BC,AB=CD,P为BC上一点,PM⊥AB于M,PN⊥CD于N,BE⊥CD于E求证:PM+PN=BE
问题描述:
如图,梯形ABCD中,AD∥BC,AB=CD,P为BC上一点,PM⊥AB于M,PN⊥CD于N,BE⊥CD于E求证:PM+PN=BE
答
∵AD∥BC,AB=CD∴四边形ABCD是等腰梯形∴∠ABC=∠DCB∵PM⊥AB,PN⊥CD,BE⊥CD∴RtΔBMP∽RtΔCEB∽RtΔCNP∴BP/BC=PM/BE,CP/BC=PN/BE上两式相加,得(BP+CP)/BC=(PM+PN)/BE→(PM+PN)/BE=1(因为BP+CP=BC)∴PM+PN=BE...