Question

已知AD、BE是三角形ABC的高,A'D'、B'E'是三角形A'B'C'的高,且BD/AD=B'D'/A'D',角C=角

Answer 1

证明:BD/AD=B'D'/A'D',又∠BDA=∠B'D'A'=90°.则⊿BDA∽⊿B'D'A',∠ABD=∠A'B'D'.

又∠C=∠C'.故⊿ABC∽⊿A'B'C',得:BC/B'C'=AD/A'D'=BE/B'E',AD·B'E'=A'D'·BE.