a,$\widehat{x'At}$=$\widehat{BAx}$=$70^o$(đối đỉnh)
⇒$\widehat{BAx}$+$\widehat{ABy}$=$70^o$+$110^o$=$180^o$(TCP)
⇒x'x//By
b,$\widehat{CBy}$=$360^o$-$110^o$-$110^o$=$140^o$
⇒$\widehat{CBy}$+$\widehat{BCz}$=$140^o$+$40^o$=$180^o$(TCP)
⇒By//Cz
CHO MK CTLHN NHA ^-^