Question

In the given figure, $\angle 3$ and $\angle 4$ are exterior angles of quadrilateral $ABCD$ at point $D$ and $B$ respectively, and $\angle A=\angle 2,\angle C=\angle 1$, prove that $\angle 3+\angle 4=\angle 1+\angle 2$

Answer 1

Let us join $A$ and $C$.

We know that, exterior angle of a triangle is equal to sum of two opposite interior angles of that triangle.

Now, $\angle 4$ is the exterior angle of $△ABC$

So,

$\angle 4=\angle ACB+\angle CAB.........\left(1\right)$

Again, $\angle 3$ is the exterior angle of $△ADC$

So,

$\angle 3=\angle ACD+\angle CAD.........\left(2\right)$

Adding eq$\left(1\right)$ and eq$\left(2\right)$

$\angle 4+\angle 3=\angle ACB+\angle CAB+\angle ACD+\angle CAD$

$\Rightarrow \angle 3+\angle 4=(\angle ACB+\angle ACD)+(\angle CAB+\angle CAD)$

$\Rightarrow \angle 3+\angle 4=\angle 1+\angle 2$

$Henceproved$ .