Question

Two sides $DA$ and $BC$ of quadrilateral $ABCD$ are produced as shown in the given figure. Show that $m+n=x+y$

Answer 1

Now,

$\angle DAB+m^0={180}^0$ (linear pair)

$\Rightarrow \angle DAB={180}^0-m^0⟶\left(1\right)$

Again,

$\angle CDA+n^0={180}^0$

$\Rightarrow \angle CDA={180}^0-n^0⟶\left(2\right)$

Also,

$\angle CDA+\angle DAB+x^0+y^0={360}^0$ (Sum of angles of a quadrilateral)

$\Rightarrow {180}^0-n^0+{180}^0-m^0+x^0+y^0={360}^0$

$\Rightarrow x^0+y^0-(m^0+n^0)={360}^0-{180}^0-{180}^0$

$\Rightarrow x^0+y^0=m^0+n^0$

Hence, proved