如图,点D在AC上,点E在CB的延长线上,且BE=AD,ED交AB于点F,求证:EF•BC=AC•FD.
问题描述:
如图,点D在AC上,点E在CB的延长线上,且BE=AD,ED交AB于点F,求证:EF•BC=AC•FD.
答
证明:过点D作DK∥BC,交AB于点K,
∴△AKD∽△ABC,△DKF∽△EBF,
∴
=DK BC
,AD AC
=DK BE
,DF EF
∴
=DK AD
,BC AC
∵BE=AD,
∴
=BC AC
,DF EF
∴EF•BC=AC•FD.