Question

如图,已知圆o是Rt三角形abc的内切圆,斜边ab与圆o相切于点d,ao的延长线交bc于点e.求证：ad×ae=ao×a

Answer 1

已知,斜边ab与圆o相切于点d,

可得：od⊥ab,

而且,ac⊥bc,∠bae = ∠cae ,

可得：ad/ao = cos∠bae = cos∠cae = ac/ae ,

所以,ad×ae = ao×ac .