求证两道几何证明题（要步骤）
1。
△ABC,△A'B'C'中,AB=A'B',BC=B'C',BD平分∠ABC,B'D'平分∠A'B'C',BD=B'D'
求证:△ABC≌△A'B'C'
简证:过A作AE∥BC交BD延长线于E,过A'作A'E'∥B'C'交B'D'延长线于E'
易证AE=AB,A'E'=A'B'==>AE=A'E'
BC/AE=BD/DE==>BC/(BC+AE)=BD/(BD+DE)=BD/BE
同理:B'C'/(B'C'+A'E')=B'D'/B'E'
BC=B'C',AE=A'E'==>B=BC/(BC+AE)=B'C'/(B'C'+A'E')
==>BD/BE=B'D'/B'E'
BD=B'D...全部
1。
△ABC,△A'B'C'中,AB=A'B',BC=B'C',BD平分∠ABC,B'D'平分∠A'B'C',BD=B'D'
求证:△ABC≌△A'B'C'
简证:过A作AE∥BC交BD延长线于E,过A'作A'E'∥B'C'交B'D'延长线于E'
易证AE=AB,A'E'=A'B'==>AE=A'E'
BC/AE=BD/DE==>BC/(BC+AE)=BD/(BD+DE)=BD/BE
同理:B'C'/(B'C'+A'E')=B'D'/B'E'
BC=B'C',AE=A'E'==>B=BC/(BC+AE)=B'C'/(B'C'+A'E')
==>BD/BE=B'D'/B'E'
BD=B'D'
==>BE=B'E'
==>△ABE≌△A'B'E'==>∠ABE=∠A'B'E'==>∠ABC=∠A'B'C'
==>△ABC≌△A'B'C'
2。
△ABC,∠B,∠C平分线BD=CE
求证:AB=AC
证明:假设AB≠AC,不妨设AB>AC
则:∠ABC∠2∠BEC>∠BDC。。。。(1)
BE/AE=BC/AC==>BE/AB=BC/(BC+AC)
CD/AD=BC/AB==>CD/AC=BC/(BC+AB)
AB>AC==>BC+ACBE/AB>CD/AC==>BE>CD
如图作平行四边形BDCF,连EF
CF=BD=CE==>∠3=∠4
BE>CD=BF==>∠6>∠5
==>∠4+∠6>∠3+∠5==>∠BEC 。。。。(2)
(1)(2)矛盾,假设不成立
所以AB=AC。收起
