Question

如圖2-5-13,已知AT切⊙O于T,ADB是割線(xiàn),BC是直徑,在AB上截取AE=AT,過(guò)E作AB的垂線(xiàn)EF,交AC延長(zhǎng)線(xiàn)于F.

求證:AB·AC=AE·AF.

圖2-5-13

Answer 1

證明:連結(jié)CD,

由切割線(xiàn)定理得AT²=AD·AB,

∵AE=AT,∴AE²=AD·AB.

∴.①

∵BC是直徑,∴∠BDC=90°,即CD⊥AB.

又EF⊥AB,∴CD∥EF.∴.②

由①②得

.∴AB·AC=AE·AF.