Question

Answer 1

Kẻ EK//AB

mà AB//CD

`=>` CD//EK (quan hệ...)

\(\Rightarrow\widehat{C}+\widehat{CEK}=180^0\left(trong-cùng-phía\right)\)

\(\Rightarrow110^0+\widehat{CEK}=180^0\)

\(\Rightarrow\widehat{CEK}=70^0\)

Có: EK//AB

\(\Rightarrow\widehat{A}+\widehat{AEK}=180^0\left(trong-cùng-phía\right)\)

\(\Rightarrow60^0+\widehat{AEK}=180^0\)

\(\Rightarrow\widehat{AEK}=120^0\)

Lại có: \(\widehat{AEK}-\widehat{CEK}=120^0-70^0=50^0=\widehat{AEC}\)