a; Ax//Dy
=>\(\hat{xAD}+\hat{ADy}=180^0\)
=>\(\hat{xAC}=180^0-\hat{CDB}\)
Ta có; \(\hat{ACB}+\hat{BCD}=180^0\) (hai góc kề bù)
=>\(\hat{ACB}=180^0-\hat{BCD}\)
Ta có: \(\hat{CBy}+\hat{CBD}=180^0\) (hai góc kề bù)
=>\(\hat{CBy}=180^0-\hat{CBD}\)
\(\hat{xAC}+\hat{ACB}+\hat{CBy}\)
\(=180^0-\hat{CDB}+180^0-\hat{BCD}+180^0-\hat{CBD}\)
\(=540^0-180^0=360^0\)
b: \(\hat{xAD}+\hat{CDB}=180^0\)
=>\(\hat{CDB}=180^0-110^0=70^0\)
\(\hat{yBC}-\hat{ACB}=30^0\)
=>\(180^0-\hat{CBD}-\left(180^0-\hat{BCD}\right)=30^0\)
=>\(\hat{BCD}-\hat{CBD}=30^0\)
Xét ΔBCD có \(\hat{BCD}+\hat{CBD}+\hat{CDB}=180^0\)
=>\(\hat{BCD}+\hat{CBD}=180^0-70^0=110^0\)
mà \(\hat{BCD}-\hat{CBD}=30^0\)
nên \(\hat{BCD}=\frac{110^0+30^0}{2}=70^0\) ; \(\hat{CBD}=70^0-30^0=40^0\)
