在△ABC(AC>AB)的边AB、AC上分别取点E、D,使BE=CD,连接ED并延长交BC的延长线于点F.求证:AB/AC=FD/EF.
问题描述:
在△ABC(AC>AB)的边AB、AC上分别取点E、D,使BE=CD,连接ED并延长交BC的延长线于点F.求证:AB/AC=FD/EF.
答
证明:过点D作DH∥AB,交CB于点H.
∵DH∥AB,
∴△DHC∽△ABC,△DHF∽△EBF.
∴DH/CD=AB/AC,DH/EB=FD/FE,.
∵BE=CD,
∴DH/EB=DH/CD.
∴AB/AC=FD/EF.