题目内容
如图,在平行四边形ABCD中,E为BC边上一点,且AB=AE.
(1)求证:△ABC≌△EAD;
(2)若AE平分∠DAB,∠EAC=25°,求∠AED的度数.

(1)求证:△ABC≌△EAD;
(2)若AE平分∠DAB,∠EAC=25°,求∠AED的度数.

(1)证明:∵四边形ABCD为平行四边形,
∴AD∥BC,AD=BC.
∴∠DAE=∠AEB.
∵AB=AE,
∴∠AEB=∠B.
∴∠B=∠DAE.
∵在△ABC和△AED中,
,
∴△ABC≌△EAD.
(2)∵AE平分∠DAB(已知),
∴∠DAE=∠BAE;
又∵∠DAE=∠AEB,
∴∠BAE=∠AEB=∠B.
∴△ABE为等边三角形.
∴∠BAE=60°.
∵∠EAC=25°,
∴∠BAC=85°.
∵△ABC≌△EAD,
∴∠AED=∠BAC=85°.
∴AD∥BC,AD=BC.

∴∠DAE=∠AEB.
∵AB=AE,
∴∠AEB=∠B.
∴∠B=∠DAE.
∵在△ABC和△AED中,
|
∴△ABC≌△EAD.
(2)∵AE平分∠DAB(已知),
∴∠DAE=∠BAE;
又∵∠DAE=∠AEB,
∴∠BAE=∠AEB=∠B.
∴△ABE为等边三角形.
∴∠BAE=60°.
∵∠EAC=25°,
∴∠BAC=85°.
∵△ABC≌△EAD,
∴∠AED=∠BAC=85°.

练习册系列答案
相关题目