问题描述：
一道超难的几何题
问题解答：
我来补答
用反证法（也可说是同一法）.
∵∠A=∠A,AD=AE
∴要证△ADC≌△AEB,只需证AB=AC
假设AB≠AC,那么AB<AC或AB>AC
当AB<AC时
在AC上取一点C'使AB=AC',显然C'在线段EC上.连结DC'交OE于O',连结CO'
∵∠A=∠A,AD=AE,AC'=AB
∴△ADC‘≌△AEB
∴∠AC'D=∠ABE
再加上∠DO'B=∠EO'C’,DB=AB-AD=AC'-AE=EC'
∴△DO'B≌△EO'C’
∴O'B=O'C'
∴O'B-OB=O'C'-OC,即O'O=O'C'-OC
∴O'O+OC=O'C'
但∠AED是锐角
∴∠DEC是钝角,∠O'C'C>∠DEC也是钝角
∴∠O'C'C>∠O'CC'
∴O'O+OC>O'C>O'C',矛盾!
同理,当AB>AC也能推出矛盾.
∴假设不成立,AB=AC
∴△ADC≌△AEB
∵∠A=∠A,AD=AE
∴要证△ADC≌△AEB,只需证AB=AC
假设AB≠AC,那么AB<AC或AB>AC
当AB<AC时
在AC上取一点C'使AB=AC',显然C'在线段EC上.连结DC'交OE于O',连结CO'
∵∠A=∠A,AD=AE,AC'=AB
∴△ADC‘≌△AEB
∴∠AC'D=∠ABE
再加上∠DO'B=∠EO'C’,DB=AB-AD=AC'-AE=EC'
∴△DO'B≌△EO'C’
∴O'B=O'C'
∴O'B-OB=O'C'-OC,即O'O=O'C'-OC
∴O'O+OC=O'C'
但∠AED是锐角
∴∠DEC是钝角,∠O'C'C>∠DEC也是钝角
∴∠O'C'C>∠O'CC'
∴O'O+OC>O'C>O'C',矛盾!
同理,当AB>AC也能推出矛盾.
∴假设不成立,AB=AC
∴△ADC≌△AEB
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