如图,等腰△ABC中,AB=AC,点D是AC上一动点,点E在BD的延长线上,且AB=AE,AF平分∠CAE交DE于F.
(1)如图1,连CF,求证:∠ABE=∠ACF;
(2)如图2,当∠ABC=60゜时,求证:AF+EF=FB;
(3)如图3,当∠ABC=45゜时,若BD平分∠ABC,求证:BD=2EF.
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