A standard geometry proof, not tied to a specific exam paper.
Problem
Prove that the sum of the interior angles of any quadrilateral is $360°$.
Key insight. A quadrilateral isn’t a shape with a directly known angle sum — but a triangle is (always $180°$). Drawing a single diagonal splits any quadrilateral into exactly two triangles, and since the diagonal’s two halves exactly reconstruct all four original interior angles between them, the quadrilateral’s angle sum is simply twice a triangle’s.
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Approach
Draw one diagonal of the quadrilateral, splitting it into two triangles. Use the known fact that the angles of any triangle sum to $180°$ for each of these two triangles. Add the two results together, and observe that this sum accounts for exactly the four original interior angles of the quadrilateral, with nothing left over or double-counted.
Solution
Step 1 — Set up the quadrilateral and draw a diagonal
Let $ABCD$ be a quadrilateral. Draw the diagonal $AC$, splitting it into $\triangle ABC$ and $\triangle ACD$.
Step 2 — Apply the angle sum of a triangle to each part
$$\angle A_1+\angle B+\angle C_1 = 180° \quad \text{(in } \triangle ABC\text{)}$$
$$\angle A_2+\angle D+\angle C_2 = 180° \quad \text{(in } \triangle ACD\text{)}$$
where $\angle A_1,\angle A_2$ are the two parts of $\angle A$ split by the diagonal, and similarly $\angle C_1,\angle C_2$ are the two parts of $\angle C$.
Step 3 — Add both equations
$$(\angle A_1+\angle A_2) + \angle B + (\angle C_1+\angle C_2) + \angle D = 180°+180° = 360°$$
Step 4 — Recognise the split angles recombine to the original angles
Since $\angle A_1+\angle A_2=\angle A$ (the diagonal splits $\angle A$ into exactly these two parts, with nothing left outside) and similarly $\angle C_1+\angle C_2=\angle C$:
$$\angle A+\angle B+\angle C+\angle D = 360°$$
Answer
The sum of the interior angles of any quadrilateral is $360°$.
Common mistakes
- Drawing the diagonal so it doesn’t pass through the interior of the quadrilateral. For a non-convex (concave) quadrilateral, one diagonal may lie outside the shape — the other diagonal must be chosen instead so that it genuinely splits the quadrilateral into two triangles covering the whole interior.
- Forgetting that $\angle A$ and $\angle C$ (the vertices the diagonal passes through) each split into two parts. Only $\angle B$ and $\angle D$ remain whole in the two triangles; treating all four angles as appearing unsplit would double-count or miscount the total.
Practise next
- Using the same diagonal-splitting method, prove that the sum of the interior angles of a pentagon is $540°$, by splitting it into three triangles from a single vertex.
Show answer
$540^\circ$. Pick any one vertex of the pentagon and draw the two diagonals from it to the non-adjacent vertices.
Those two diagonals cut the pentagon into three triangles, and every interior angle of the pentagon is accounted for exactly once across them — the angles at the chosen vertex are split between the triangles, and the rest sit whole inside one triangle each.
So the total is $3\times180^\circ=540^\circ$. The same count gives $(n-2)\times 180^\circ$ in general, since an $n$-gon splits into $n-2$ triangles from one vertex.
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