Question

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

Answer 1

Here,
$\angle BAD=180°-b$ [linear pair angles]
Similarly,
$\angle BCD=180°-a$ [linear pair angles]
We know that, sum of all interior angles of a quadrilateral is equal to $360°$.
Therefore,
$\angle BAD+\angle ABC+\angle BCD+\angle ADC=360°$
$180°-b+x+180°-a+y=360°$
$360°+x+y-(a+b)=360°$
$x+y-(a+b)=0$
$x+y=a+b$
Hence Proved.