Question

The sides $AD$ and $BC$ of a quadrilateral are produced as shown in the given figure. Prove that $x=\frac{a+b}{2}$.

Answer 1

We have, $\angle ADC={180}^{\circ }-a$

And $\angle BCD={180}^{\circ }-b$

As we know sum of angle in the quadrilateral will be ${360}^{\circ }$4

So, $x+x+{180}^{\circ }-a+{180}^{\circ }-b=360$\circ$

$⟹2x-a-b=0$

$⟹x=\frac{a+b}{2}$