已知BE是等腰三角形ABC的角平分线,∠ACB=90°,延长BC到点D,使CD=CE,连结AD与BE的延长线交于点F.

问题描述:

已知BE是等腰三角形ABC的角平分线,∠ACB=90°,延长BC到点D,使CD=CE,连结AD与BE的延长线交于点F.
证明AE*AC=2AF^2

AC=BC,CD=CE
∠ACB=∠ACd=90°
△CBE≌△ACD
∠DAC=∠CBE
∠DAC+∠D=90°
∠CbE +∠D =90°
BF⊥AD
BE是角平分线
AF=FD
△AFE∽△ACD
AF:AC=AE:AD=AE:2AF
AE*AC=2AF^2