解题思路:
试题分析
:
(
1)
∵
∠
1
=
∠
2
,
AC
=
A
C
,
∠3
=
∠
4
∴
△
ABC
≌
△
ADC
;
(2)
∵
△
ABC
≌
△
ADC
∴
A
B
=
A
D
∵
∠
1
=
∠
2
,
AO
=
A
O
∴
△
ABO
≌
△
ADO
∴
B
O
=
D
O
.
(1)由∠1=∠2,∠3=∠4,再结合公共边AC即可作出判断;
(1)由△ABC≌△ADC证得AB=AD,再结合∠1=∠2,公共边AO可证得△ABO≌△ADO,问题得证.
解题思路:
试题分析
:
(
1)
∵
∠
1
=
∠
2
,
AC
=
A
C
,
∠3
=
∠
4
∴
△
ABC
≌
△
ADC
;
(2)
∵
△
ABC
≌
△
ADC
∴
A
B
=
A
D
∵
∠
1
=
∠
2
,
AO
=
A
O
∴
△
ABO
≌
△
ADO
∴
B
O
=
D
O
.
(1)由∠1=∠2,∠3=∠4,再结合公共边AC即可作出判断;
(1)由△ABC≌△ADC证得AB=AD,再结合∠1=∠2,公共边AO可证得△ABO≌△ADO,问题得证.