Bài 3:
Kẻ $Ba$ về phía bên trái ao cho $Ba//Ax$
$→\widehat{ABa}+\widehat{BAx}=180^\circ$ (trong cùng phía)
mà $\widehat{BAx}=150^\circ$
$→\widehat{ABa}=180^\circ-150^\circ=30^\circ$
$→\widehat{CBa}=70^\circ-30^\circ=40^\circ$
$→\widehat{CBa}+\widehat{BCy}=40^\circ+140^\circ=180^\circ$
mà 2 góc ở vị trí trong cùng phía
$→Ba//Cy$ mà $Ba//Ax$
$→Ax//Cy$
Bài 4:
a) $\widehat{Q_1}+\widehat{Q_2}=180^\circ$
mà $\widehat{Q_1}-\widehat{Q_2}=80^\circ$
$→\widehat{Q_1}=\dfrac{180^\circ+80^\circ}{2}=130^\circ$
$→\widehat{Q_2}=\dfrac{180^\circ-80^\circ}{2}=50^\circ$
mà $m//n$
$→\widehat{Q_1}=\widehat{P_2}=130^\circ$ (so le trong)
$→\widehat{Q_2}=\widehat{P_1}=50^\circ$ (so le trong)
b) $\widehat{Q_1}+\widehat{Q_2}=180^\circ$
mà $\widehat{Q_2}=\dfrac{5}{13}\widehat{Q_1}$
$→\widehat{Q_1}+\dfrac{5}{13}\widehat{Q_1}=180^\circ$
$→\dfrac{18}{13}\widehat{Q_1}=180^\circ$
$→\widehat{Q_1}=130^\circ$
$→\widehat{Q_2}=50^\circ$
mà $m//n$
$→\widehat{Q_1}=\widehat{P_2}=130^\circ$ (so le trong)
$→\widehat{Q_2}=\widehat{P_1}=50^\circ$ (so le trong)
