a//b
a⊥d
Do đó: b⊥d
=>\(\hat{B}=90^0\)
Ta có: \(\hat{D_1}+\hat{D_2}=180^0\) (hai góc kề bù)
=>\(\hat{D_1}=180^0-45^0=135^0\)
a//b
=>\(\hat{D_2}=\hat{E_1}\) (hai góc đồng vị)
=>\(\hat{E_1}=45^0\)
Ta có: b//c
=>\(\hat{E_1}+\hat{G_1}=180^0\) (hai góc ngoài cùng phía)
=>\(\hat{G_1}=180^0-45^0=135^0\)
