Question

Answer 1

Lời giải:
$\widehat{B_3}=\widehat{B_1}=70^0$ (hai góc đối đỉnh)

$\widehat{B_2}+\widehat{B_1}=180^0$

$\widehat{B_2}=180^0-\widehat{B_1}=180^0-70^0=110^0$

$\widehat{B_4}=\widehat{B_2}=110^0$

$\widehat{C_3}=\widehat{B_1}=70^0$ (hai góc đồng vị)

$\widehat{C_1}=\widehat{C_3}=70^0$ (hai góc đối đỉnh)

$\widehat{C_4}+\widehat{C_3}=180^0$

$\widehat{C_4}=180^0-\widehat{C_3}=180^0-70^0=110^0$

$\widehat{C_2}=\widehat{C_4}=110^0$ (hai góc đối đỉnh)