Question

The sides $BA$ and $DC$ of a quadrilateral $ABCD$ are produced as shown in the figure. Prove that $a+b=x+y$

Answer 1

From, figure

$b^{\circ }+\angle DAB={180}^{\circ }$

$⟹\angle DAB={180}^{\circ }-b^{\circ }$.......(i)

$a^{\circ }+\angle BCD={180}^{\circ }$

$⟹\angle BCD={180}^{\circ }-a^{\circ }$.......(ii)

and $\angle A+\angle B+\angle C+\angle D={360}^{\circ }$

$⟹180-b+x+180-a+y=360$

$⟹x+y=360-180+b-180+a$

$\therefore x+y=a+b$

Hence proved.