Question

直角梯形ABCD中,AD平行于BC,∠B=90°,且AD+BC=CD,以AB为直径作圆O,求证 CD与圆0相切

Answer 1

用面积做

过O做OE垂直于CD 连结OD OC

梯形ABCD面积 = 三角形OAD+OBC+ODC

= 1/2*AO*AD+1/2*BO*BC+1/2*OE*CD

= 1/2*AO*(AD+BC)+1/2*OE*CD

= 1/2*(AO+EO)*CD

梯形ABCD面积 = 1/2*AB*(AD+BC) = 1/2*2AO*CD = AO*CD

所以1/2*(AO+EO)*CD = AO*CD

AO+EO=2AO

EO=AO

则圆点O到CD的距离为OE,CD与圆O相切于E