a, Vì a;b phân biệt mà \(a\perp CD;b\perp CD\) nên
\(a\text{//}b\)(do hai đường thẳng phần biệt cùng vuông góc với đường thẳng thứ 3 thì chúng song song với nhau)(đpcm)
b, Vì a//b(cmt) nên \(\widehat{K_1}=\widehat{H_2}\left(slt\right);\widehat{H_1}=\widehat{K_2}\left(slt\right);\widehat{H_2}+\widehat{K_2}=180^o\left(tcp\right)\)(1)
mà \(\widehat{K_1}=2\widehat{H_1}\)(gt)
\(\Rightarrow\widehat{H_2}=2\widehat{K_2}\)(2)
Thay (2) vào (1) ta có:
\(2\widehat{K_2}+\widehat{K_2}=180^o\Rightarrow3\widehat{K_2}=180^o\)
\(\Rightarrow\widehat{K_2}=60^o\)
\(\Rightarrow\widehat{H_2}=180^o-\widehat{K_2}=180^o-60^o=120^o\)
Vậy..................
a) Vì 
\(\left\{{}\begin{matrix}\widehat{C_2}=90^o&\widehat{D_2}=90^o&\end{matrix}\right.\)
\(\Rightarrow\widehat{C_2}=\widehat{D_2}\) (2 góc đồng vị)
\(\Rightarrow\)a//b\(\left(ĐPCM\right)\)
b) Vì a//b (cmt)
\(\Rightarrow\widehat{K_1}+\widehat{H_1}=180^o\) (2 góc trong cùng phía)
Mà \(\widehat{K_1}=2\widehat{H_1}\) (gt)
\(\Rightarrow2\widehat{H_1}+\widehat{H_1}=180^o\)
\(\Rightarrow3\widehat{H_1}=180^o\)
\(\Rightarrow\widehat{H_1}=\dfrac{180}{3}\)
\(\Rightarrow\widehat{H_1}=60^o\)
\(\Rightarrow\widehat{K_1}=2.60=120^o\)
Vậy \(\left\{{}\begin{matrix}\widehat{H_1}=60^o\\\widehat{K_1}=120^o\end{matrix}\right.\)
nhục méo bt lm xl bn
