91Ó°ÊÓ

The theorem for the sum of interior angles of a polygon states that the sum of measures of the interior angles of a polygon with sides is $\left(n-2\right)180$.

The number of sides in a quadrilateral is 4.

Therefore, the value of $n$is 4.

Therefore, the sum of interior angles of a quadrilateral is :

$$\begin{gathered} \left(4−2\right)180=2×180 \\ =360 \end{gathered}$$

Therefore, the sum of interior angles of a quadrilateral is $360°$.

As, the sum of interior angles of a quadrilateral is $360°$.

$$\begin{gathered} m∠A+m∠B+m∠C+m∠D=360 \\ x+2x+3x+4x=360 \\ 10x=360 \\ x=\frac{360}{10} \\ x=36 \end{gathered}$$

Therefore, the value of x is 36.

As, the value of x is 36.

Therefore,

$m∠A=x=36°$

$m∠B=2x=2×36=72°$

$m∠C=3x=3×36=108°$

$m∠D=4x=4×36=144°$

It can be noticed that:

$m∠A+m∠B=36°+144°=180°$

$m∠C+m∠D=108°+72°=180°$

Therefore, the pair of sides are AD and BC must be parallel.