如圖,在△ABC中,AB=AC,點(diǎn)D、E、F分別在AB、BC、CA邊上,且BD=CE,∠DEF=∠ABC.
![](http://hiphotos.baidu.com/zhidao/pic/item/b17eca8065380cd720c4225ca244ad34588281ad.jpg)
(1)求證:△EDB≌△FEC.
(2)若點(diǎn)D、E、F分別在AB、BC、CA邊或它們某一方的延長(zhǎng)線上(至少一個(gè)點(diǎn)在延長(zhǎng)線上),其他條件不變,畫出一種符合題意的圖形,并要求且說明此時(shí)(1)中的結(jié)論仍成立.
(1)∵∠BDE+∠BED=180°-∠ABC,∠BED+∠FEC=180°-∠DEF,
又∠DEF=∠ABC,∴∠BDE+∠BED=∠BED+∠FEC,即∠BDE=∠FEC,
∵AB=AC,∴∠B=∠C,又BD=CE,
∴△EDB≌△FEC;
![](http://hiphotos.baidu.com/zhidao/pic/item/2fdda3cc7cd98d10c30e8c80223fb80e7aec90ad.jpg)
(2)根據(jù)題意畫出圖形,如圖所示:
∵∠ABC=∠BDE+∠BED,∠DEF=∠CEF+∠BED,且∠DEF=∠ABC,
∴∠BDE+∠BED=∠CEF+∠BED,即∠BDE=∠CEF,
∵AB=AC,∴∠ABC=∠ACB,∴∠DBE=∠ECF,又BD=CE,
∴△EDB≌△FEC.