Angles and Polygons

The cut-and-rearrange activity suggests the result; here we prove it. Through the vertex \(C\), draw the line parallel to the opposite side \(\Line{AB}\), meeting it nowhere. The two sides \(\Segment{CA}\) and \(\Segment{CB}\) are then transversals cutting a pair of parallel lines. Each base angle reappears at \(C\) as an alternate angle: the transversal \(\Segment{CA}\) gives \(\Angle{LCA}=c^\circ\), and the transversal \(\Segment{CB}\) gives \(\Angle{BCR}=b^\circ\). The three angles at \(C\) lie along the straight line \(\Line{LR}\), so they form a straight angle: $$\begin{aligned}[t] \Angle{LCA}+\Angle{ACB}+\Angle{BCR} &= 180^\circ && \text{(angles on a straight line)}\\ c^\circ + a^\circ + b^\circ &= 180^\circ. && \text{(alternate angles, parallel lines)} \end{aligned}$$ Hence the three interior angles satisfy \(a^\circ+b^\circ+c^\circ=180^\circ\).

We cut the quadrilateral \(ABCD\) into two triangles, \(ABC\) and \(ACD\), along the diagonal \(\Segment{AC}\), which lies inside it (for a concave quadrilateral, the diagonal to choose is given in the Remark “Angle Sum of a Concave Polygon”). $$\begin{aligned} \text{Sum of the interior angles of quadrilateral } ABCD &= \text{Sum of the interior angles of } \triangle ABC \\ &\quad + \text{Sum of the interior angles of } \triangle ACD \\ &= 180^\circ + 180^\circ \\ &= 360^\circ \end{aligned}$$

We prove it for a convex polygon \(A_1A_2 \ldots A_n\); the Remark below deals with a concave one. From the vertex \(A_1\), draw the diagonals to every vertex except \(A_1\) and its two neighbours \(A_2\) and \(A_n\): there are \(n - 3\) of them. As the polygon is convex, they lie inside it and cut it into the \(n - 2\) triangles \(A_1A_2A_3\), \(A_1A_3A_4\), …, \(A_1A_{n-1}A_n\) — one triangle more than diagonals. Here is the hexagon: \(3\) diagonals, \(4\) triangles. The angles of these triangles fill the interior angles of the polygon exactly: at \(A_1\), the angles of all the triangles together make the angle of the polygon, and at every other vertex, one or two triangles fill it. Each triangle has an angle sum of \(180^\circ\) (Proposition “Sum of the angles of a triangle”), so $$\begin{aligned}[t] \text{sum of the angles of the polygon} &= \text{sum of the angles of the } n - 2 \text{ triangles} && \text{(the triangles fill the angles)}\\ &= (n - 2) \times 180^\circ. \end{aligned}$$

Let the convex polygon have \(n\) sides. At each of its \(n\) vertices, the interior angle and the exterior angle add up to \(180^\circ\), so all the interior and exterior angles together make \(n \times 180^\circ\). The interior angles alone make \((n - 2) \times 180^\circ\) (Proposition “Angle Sum of a Polygon”). Hence $$\begin{aligned}[t] \text{sum of the exterior angles} &= n \times 180^\circ - (n - 2) \times 180^\circ && \text{(all the angles minus the interior angles)}\\ &= n \times 180^\circ - n \times 180^\circ + 2 \times 180^\circ && \text{(distributive law)}\\ &= 360^\circ. \end{aligned}$$

A regular polygon has all its angles equal (Definition “Regular Polygon”). Its \(n\) equal interior angles add up to \((n - 2) \times 180^\circ\), so each of them measures \(\dfrac{(n - 2) \times 180^\circ}{n}\); each is less than \(180^\circ\), so the polygon is convex. At each vertex, the exterior angle is \(180^\circ\) minus the interior angle, so the \(n\) exterior angles are equal too; they add up to \(360^\circ\) (Proposition “Sum of the Exterior Angles”), so each of them measures \(\dfrac{360^\circ}{n}\). Finally, the two forms of the interior angle agree: $$\begin{aligned}[t] \frac{(n - 2) \times 180^\circ}{n} &= \frac{n \times 180^\circ - 360^\circ}{n} && \text{(distributive law)}\\ &= 180^\circ - \frac{360^\circ}{n}. \end{aligned}$$