AE = 2CD .证明如下:
延长AB、CD,相交于点F.
在△ACD和△AFD中,∠ADC = 90°= ∠ADF ,AD为公共边,∠CAD = ∠FAD ,
所以,△ACD ≌ △AFD ,可得:CD = DF ,即有:CF = 2CD ;
在△CBF和△ABE中,∠BCF = 90°-∠BFC = ∠BAE ,CB = AB ,∠CBF = 90°= ∠ABE ,
所以,△CBF ≌ △ABE ,可得:CF = AE ,则有:AE = 2CD .
AE = 2CD .证明如下:
延长AB、CD,相交于点F.
在△ACD和△AFD中,∠ADC = 90°= ∠ADF ,AD为公共边,∠CAD = ∠FAD ,
所以,△ACD ≌ △AFD ,可得:CD = DF ,即有:CF = 2CD ;
在△CBF和△ABE中,∠BCF = 90°-∠BFC = ∠BAE ,CB = AB ,∠CBF = 90°= ∠ABE ,
所以,△CBF ≌ △ABE ,可得:CF = AE ,则有:AE = 2CD .