(1)
已知,Rt△ABC中,∠BAC = 90°,AB = AC ,
可得:∠ABC = 45°;
所以,∠AMD = ∠BMH = 90°-∠CBD = 90°-(1/2)∠ABC = 67.5°.
(2)
在△BCE和△BFE中,
∠BEC = 90°= ∠BEF ,BE为公共边,∠CBE = ∠FBE ,
所以,△BCE ≌ △BFE ,
可得:CE = EF ,即有:CF = 2CE ;
在△CAF和△BAD中,
∠ACF = 90°-∠AFC = ∠ABD ,AC = AB ,∠CAF = 90°= ∠BAD ,
所以,△CAF ≌ △BAD ,
可得:CF = BD ,则有:BD = 2CE .