In quadrilateral $ABC\text{}D,m\angle A=x.m\angle B=2x,m\angle C=3x,$and $m\angle D=4x.$. Find the value of and then state which pair of sides of ABCD must be parallel.

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 :

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

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

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

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

Therefore, the value of x is 36.

As, the value of x is 36.

Therefore,

$m\angle A=x=36°$

$m\angle B=2x=2\times 36=72°$

$m\angle C=3x=3\times 36=108°$

$m\angle D=4x=4\times 36=144°$

It can be noticed that:

$m\angle A+m\angle B=36°+144°=180°$

$m\angle C+m\angle D=108°+72°=180°$

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