In today’s lesson, we will prove that the sum of interior angles in a simple convex quadrangle is always 360° by showing that this quadrangle is composed of two triangles.

## Problem

Show that the sum of interior angles in a simple convex quadrangle is always 360°.

## Strategy

Let’s think about the strategy to answer this geometry problem. 360 is a familiar number – it is the measure of angles in a circle, but there are no circles here, so let’s rule that out.

360 is also 2×180, and that is also a familiar number. It is the sum of the angles in a triangle.

So, if we can show that this quadrangle is composed of two triangles, we will have the proof- and that’s all we need.

## Proof

In any simple convex polygon, a line connecting 2 points on the perimeter of the polygon is entirely within the polygon, per the definition of convex polygons.

So, let’s draw a line connecting two opposite corners of the quadrangle (such a line is a called a ‘diagonal’) – say from C to A:

(1) m∠A_{1} + m∠D+ m∠C_{1} = 180° //sum of the interior angles in triangle ΔADC

(2) m∠A_{2} + m∠B+ m∠C_{2} = 180° //sum of the interior angles in triangle ΔABC

(3) m∠A_{1} + m∠D+ m∠C_{1}+m∠A_{2} + m∠B+ m∠C_{2 }= 360° //add both equations

(4) m∠A_{1} + m∠A_{2} +m∠D+ m∠C_{1}+ m∠C_{2} + m∠B= 360° //re-arrange terms

(5) m∠A_{1} + m∠A_{2 =} m∠A // angle addition postulate

(6) m∠C_{1} + m∠C_{2 =} m∠C // angle addition postulate

(7) m∠A+m∠D+ m∠C + m∠B= 360° ∎ //substitution property of equality