KA)BAD = BHD  BA = BH B) KẺ BM  EK BM = AE = AB = BH BMK = BHK...

Bài3:

K

a)BAD = BHD  BA = BH

b) Kẻ BM  EK BM = AE = AB = BH

BMK = BHK góc B

3

= góc B

4

H

Mà, góc B

1

= góc B

2

AEC

góc B

1

+ góc B

2

+ góc B

3

+ góc B

4

= 90

0

D góc DBK = góc B

2

+ góc B

3

= 45

0