Interior angle sum of a quadrilateral and an n-gon

Step 1 (split the quadrilateral along a diagonal). In convex quadrilateral $ABCD$, draw the diagonal $AC$. Because the figure is convex, this segment lies entirely inside the quadrilateral and partitions it into two triangles $\triangle ABC$ and $\triangle ACD$.

Step 2 (apply the triangle angle sum to each piece). In $\triangle ABC$ we have $\angle BAC + \angle B + \angle BCA = 180^\circ$, and in $\triangle ACD$ we have $\angle CAD + \angle D + \angle DCA = 180^\circ$.

Step 3 (add the two equations and generalize). Adding both equations together gives $(\angle BAC + \angle CAD) + \angle B + (\angle BCA + \angle DCA) + \angle D = 180^\circ + 180^\circ = 360^\circ$. Since the adjacent angles at the two ends of the diagonal recombine into the full vertex angles $\angle BAC + \angle CAD = \angle A$ and $\angle BCA + \angle DCA = \angle C$, this simplifies directly to $\angle A + \angle B + \angle C + \angle D = 360^\circ$. For a convex polygon with $n$ vertices, drawing all $n - 3$ diagonals from one vertex cuts the interior into $n - 2$ triangles whose angles add up to $(n - 2) \cdot 180^\circ$.