Question

In the adjoining figure $\underset{AB}{\to }$ || $\underset{DE}{\to }$. Prove that $\angle ABC-\angle DCB+\angle CDE={180}^o$

Answer 1

Construction:
Through C draw $\underset{XY}{\to }$ || $\underset{AB}{\to }$
Proof:
$\angle ABC+\angle BCX={180}^o$ [sum of interior angles of the same side of transversal]
hence $\angle BCX=180-\angle ABC$
$\angle CDE+\angle DCY-{180}^o$ [sum of interior angles of the same side of transversal]
hence $\angle DCY=180-\angle CDE$
$\angle BCX+\angle BCD+\angle DCY=180$ [XCY is straight angle]
$180-\angle ABC+\angle BCD+180-\angle CDY=180$ [by substituting]
$360-\angle ABC+\angle BCD-180-\angle CDE$
${180}^o=\angle ABC-\angle BCD+\angle CDE$
OR
$\angle ABC-\angle BCD+\angle CDE={180}^o$