已知,在正方形ABCD中,对角线AC与BD相交于点E,AF平分∠BAC,交BD于F

E1F1 + A1C1/2 = AB证： 设正方形ABCD连长为a, 如图所见的,作F1P⊥AB, F1Q⊥BC 根据角平分,以及ABCD为正方形,可知F1P = F1Q = F1E1 = PB = BQ (不妨设为都=t) 还有,CC1 = AA1 = A1O (= b) 于是,有 C1E1 = C1Q = a-b-t 故,AP = OE1 =a-t 又C1E1+A1O =a-b-t+b =a-t =OE1, 则OE1 =A1C1/2 故,E1F1 + A1C1/2 =E1F1+ OE1 =PB+AP =AB