Prove that the sum of angles of any convex quadrilateral is $360^{\circ} .$

(N/A) Let $ABCD$ be a convex quadrilateral.
Draw a diagonal $AC$ which divides the quadrilateral into two triangles,$\triangle ABC$ and $\triangle ADC$.
We know that the sum of the angles of a triangle is $180^{\circ}$.
For $\triangle ABC$,$\angle BAC + \angle ABC + \angle BCA = 180^{\circ}$ (Equation $1$).
For $\triangle ADC$,$\angle DAC + \angle ADC + \angle DCA = 180^{\circ}$ (Equation $2$).
Adding Equation $1$ and Equation $2$,we get:
$(\angle BAC + \angle DAC) + \angle ABC + \angle ADC + (\angle BCA + \angle DCA) = 180^{\circ} + 180^{\circ}$.
From the figure,$\angle BAC + \angle DAC = \angle A$ and $\angle BCA + \angle DCA = \angle C$.
Therefore,$\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.
Hence,the sum of the angles of a convex quadrilateral is $360^{\circ}$.