Question

BE,CF分别是三角形ABC的边AC、AB上的高,BE与CF相交于点D.求证：1.三角形ABC相似三角形AEF

Answer 1

1.证明:∠AEB=∠AFC=90度,∠A=∠A.则⊿AEB∽⊿AFC,AF/AE=AC/AB;

又∠EAF=∠BAC,则⊿ABC∽⊿AEF.

2.BE垂直AC,角A=60度,则∠ABE=30度,AE/AB=1/2.

⊿ABC∽⊿AEF,则S⊿AEF/S⊿ABC=(AE/AB)^2=(1/2)^2=1/4.