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

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, i.e., $m\angle DAB + m\angle B + m\angle BCD + m\angle D = 360$.

2. Solution Steps

We are given a quadrilateral $ABCD$. We can divide the quadrilateral into two triangles by drawing a diagonal, such as $AC$.

The sum of the interior angles of a triangle is 180 degrees.

$m\angle DAC + m\angle DCA + m\angle D = 180$ (Triangle $ADC$)

$m\angle BAC + m\angle BCA + m\angle B = 180$ (Triangle $ABC$)

Adding the two equations gives us:

$m\angle DAC + m\angle DCA + m\angle D + m\angle BAC + m\angle BCA + m\angle B = 180 + 180$

$m\angle DAC + m\angle BAC + m\angle DCA + m\angle BCA + m\angle D + m\angle B = 360$

Notice that $m\angle DAB = m\angle DAC + m\angle BAC$ and $m\angle BCD = m\angle BCA + m\angle DCA$.

So, substituting these values into the equation gives us:

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

Rearranging the terms gives:

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

3. Final Answer

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