The problem states that $ABCD$ is a quadrilateral. We need to prove that the sum of its interior angles is $360$ degrees. In other words, we need to prove that $m\angle DAB + m\angle B + m\angle BCD + m\angle D = 360^\circ$.

1. Problem Description

The problem states that $ABCD$ is a quadrilateral. We need to prove that the sum of its interior angles is $360$ degrees. In other words, we need to prove that $m\angle DAB + m\angle B + m\angle BCD + m\angle D = 360^\circ$.

2. Solution Steps

We are given a quadrilateral $ABCD$. The diagram shows a diagonal $AC$. This diagonal divides the quadrilateral into two triangles, $\triangle ADC$ and $\triangle ABC$.

The sum of the interior angles of a triangle is $180^\circ$.

Therefore, in $\triangle ADC$, we have

$m\angle DAC + m\angle ACD + m\angle D = 180^\circ$

And in $\triangle ABC$, we have

$m\angle BAC + m\angle ACB + m\angle B = 180^\circ$

Adding these two equations gives

$m\angle DAC + m\angle ACD + m\angle D + m\angle BAC + m\angle ACB + m\angle B = 180^\circ + 180^\circ$

$m\angle DAC + m\angle ACD + m\angle D + m\angle BAC + m\angle ACB + m\angle B = 360^\circ$

Rearranging the terms, we have

$(m\angle DAC + m\angle BAC) + m\angle B + (m\angle ACD + m\angle ACB) + m\angle D = 360^\circ$

We see that $m\angle DAB = m\angle DAC + m\angle BAC$ and $m\angle BCD = m\angle ACD + m\angle ACB$.

Substituting these, we get

$m\angle DAB + m\angle B + m\angle BCD + m\angle D = 360^\circ$

3. Final Answer

$m\angle DAB + m\angle B + m\angle BCD + m\angle D = 360^\circ$