Cho Hình 3.26, Biết Ax//Dy, $\widehat{xAC}=50^{\circ},\widehat{ACD}=110^{\circ}$

Kẻ một đường t qua C và Ct// Ax mà Ax//Dy suy ra Ct//Ax//Dy.

Ta có: Ct//Ax suy ra $\widehat{CAx}=\widehat{ACt}=50^{\circ}$ (hai góc so le trong)

Ta có: $\widehat{ACt}+\widehat{DCt}=\widehat{DCA}$

=>$ 50^{\circ}+\widehat{DCt}=110^{\circ}$

=> $\widehat{DCt}=110^{\circ}-50^{\circ}=60^{\circ}$

Vì Dy//Ct => $\widehat{DCt}=\widehat{CDy}=60^{\circ}$ (hai góc so le trong)

Vậy $\widehat{CDy}=60^{\circ}$