Geometric Reasoning and Proof

1 What a Proof Is

A proof is a chain of logical steps that starts from facts you are allowed to assume and ends at the statement you were asked to establish. Every link in the chain needs a reason.

Checking examples is not a proof. Showing that a result works for three particular triangles proves nothing about the fourth. A proof must work for every case at once — which is why it uses letters, not numbers.

What you may assume without proof at GCSE:
• the basic angle facts (straight line, point, vertically opposite)
• the parallel-line rules (alternate, corresponding, co-interior)
• the angle sum of a triangle
• the four congruence criteria (SSS, SAS, ASA, RHS)
• properties of named shapes, once you have named the shape
Everything else you must derive.

The shape of every geometric proof

Statement → reason → Statement → reason → … → Conclusion

2 The Language of Proof

Word in the question	What is expected
Prove / Show that	A full logical argument with a reason at every step.
Give a reason	Quote the standard angle fact by name.
Explain why	A sentence or two of reasoning; usually fewer marks than "prove".
Deduce	Use a result you have just proved to get something new.
Hence	You must use the previous part of the question.
Hence or otherwise	Using the previous part is the intended route, but any valid method scores.

Write reasons in the standard wording. "Alternate angles are equal" earns the mark; "the Z rule" may not. Learn the phrases: angles on a straight line, angles at a point, vertically opposite angles, angle sum of a triangle, alternate angles, corresponding angles, co-interior angles, base angles of an isosceles triangle, exterior angle of a triangle.

3 Angle Chase Proofs

The most common type: introduce a letter for an unknown angle, chase it round the diagram, and arrive at the required result.

Let the unknown angle be $x$ (and a second one $y$ if needed).
Express other angles in terms of $x$, giving a reason each time.
Form an equation from an angle sum.
Simplify to reach the required statement.

Worked Example 1 — Proving the angle sum of a triangle

Prove that the angles of a triangle add to $180^\circ$.

①Draw triangle $ABC$, and draw a line through $A$ parallel to $BC$.

②Label the three angles at $A$ (left to right) as $p$, $\angle BAC$ and $q$. These three angles lie on a straight line, so $p + \angle BAC + q = 180^\circ$.

③$p = \angle ABC$ (alternate angles, the new line is parallel to $BC$)

④$q = \angle ACB$ (alternate angles, the new line is parallel to $BC$)

⑤Substituting: $\angle ABC + \angle BAC + \angle ACB = 180^\circ$ ∎

Worked Example 2 — Proving the exterior angle rule

In triangle $ABC$, $BC$ is extended to $D$. Prove that $\angle ACD = \angle ABC + \angle BAC$.

①Let $\angle ABC = a$, $\angle BAC = b$ and $\angle ACB = c$.

②$a + b + c = 180^\circ$ (angle sum of a triangle)

③$\angle ACD + c = 180^\circ$ (angles on a straight line $BCD$)

④From ②, $c = 180 - a - b$. Substituting into ③:

⑤$\angle ACD + 180 - a - b = 180$, so $\angle ACD = a + b$ ∎

4 Proofs Using Congruence

When a question asks you to prove two lengths or two angles are equal, congruence is nearly always the tool.

The standard route

Find two triangles → prove them congruent → deduce that the parts you want are equal

Worked Example 3 — Diagonals of a rectangle

Prove that the diagonals of a rectangle are equal in length.

①Let $ABCD$ be a rectangle. Consider triangles $ABC$ and $BAD$.

②$AB = BA$ (common side)

③$\angle ABC = \angle BAD = 90^\circ$ (angles of a rectangle)

④$BC = AD$ (opposite sides of a rectangle are equal)

⑤Therefore $\triangle ABC \cong \triangle BAD$ (SAS).

⑥Corresponding sides are equal, so $AC = BD$. The diagonals are equal. ∎

Worked Example 4 — A two-stage proof

$ABCD$ is a parallelogram. $E$ is the midpoint of $AB$ and $F$ is the midpoint of $CD$. Prove that $DE = BF$.

①$AB = CD$ (opposite sides of a parallelogram), so halving gives $AE = CF$.

②$AD = CB$ (opposite sides of a parallelogram)

③$\angle DAE = \angle BCF$ (opposite angles of a parallelogram are equal)

④Therefore $\triangle DAE \cong \triangle BCF$ (SAS).

⑤Corresponding sides give $DE = BF$. ∎

5 Proving Pythagoras' Theorem

The classic dissection proof uses four copies of the same right-angled triangle arranged inside a square.

Worked Example 5 — The dissection proof

Prove that $a^2 + b^2 = c^2$ for a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$.

①Arrange four copies of the triangle inside a square of side $(a + b)$, as shown. The inner tilted quadrilateral has all sides $c$.

②At each corner of the inner shape, the two acute angles of the triangle sit either side. Those two angles add to $90^\circ$, and the three angles lie on a straight line, so each inner angle is $180 - 90 = 90^\circ$. The inner shape is a square of side $c$.

③Area of the large square $= (a+b)^2 = a^2 + 2ab + b^2$

④Area also $= 4 \times \left(\tfrac{1}{2}ab\right) + c^2 = 2ab + c^2$

⑤Equating: $a^2 + 2ab + b^2 = 2ab + c^2$

⑥Subtracting $2ab$ from both sides: $a^2 + b^2 = c^2$ ∎

6 Algebraic Proof in Geometry

Some proofs need algebra rather than a diagram argument. The method is to write everything in terms of letters and simplify.

Worked Example 6 — Interior angle sum of a polygon

Prove that the interior angles of an $n$-sided polygon add to $(n-2) \times 180^\circ$.

①Choose one vertex of the polygon and join it to every other non-adjacent vertex.

②There are $n$ vertices; the chosen one and its two neighbours cannot be joined, so there are $n - 3$ diagonals.

③These $n - 3$ diagonals cut the polygon into $n - 2$ triangles.

④Every angle of every triangle is part of an interior angle of the polygon, and nothing is left over.

⑤Each triangle contributes $180^\circ$, so the total is $(n-2) \times 180^\circ$ ∎

Worked Example 7 — Disproving a statement

Alex claims: "If a quadrilateral has two pairs of equal sides, it must be a parallelogram." Show that Alex is wrong.

①To disprove a general claim you only need one counter-example.

②A kite has two pairs of equal sides — but the equal sides are adjacent, not opposite.

③A kite is not a parallelogram, because its opposite sides are neither equal nor parallel.

So the claim is false. The correct statement is: if the two pairs of equal sides are opposite pairs, the quadrilateral is a parallelogram.

To disprove, find one counter-example. To prove, argue in general. This asymmetry is the single most useful idea about proof.

7 Quick Reference

Structure

Statement, reason, statement, reason… then the conclusion.

Examples ≠ proof

Testing cases never proves a general result.

Counter-example

One is enough to disprove a claim.

Equal lengths?

Look for two triangles and prove congruence.

Introduce letters

Let the unknown angle be $x$, then chase it round.

Add a line

A parallel line or a diagonal often unlocks the proof.

"Hence"

You must use the previous part of the question.

Standard wording

Quote angle facts by their proper names.

Finish it

End with the statement you were asked to prove, and mark it ∎.

8 Practice Questions

Question 1

$ABC$ is a straight line. $\angle DBA = 3x$ and $\angle DBC = 2x + 40$. Find $x$, giving a reason.

▶ Show solution

$3x + (2x + 40) = 180$ — angles on a straight line add to $180^\circ$.

$5x + 40 = 180 \Rightarrow 5x = 140$

$x = 28$

Question 2

Prove that the opposite angles of a parallelogram are equal.

▶ Show solution

Let $ABCD$ be a parallelogram with $AB \parallel DC$ and $AD \parallel BC$.

$\angle A + \angle B = 180^\circ$ (co-interior angles, $AD \parallel BC$)

$\angle B + \angle C = 180^\circ$ (co-interior angles, $AB \parallel DC$)

Therefore $\angle A + \angle B = \angle B + \angle C$.

Subtracting $\angle B$ from both sides gives $\angle A = \angle C$.

The same argument starting from $\angle B$ gives $\angle B = \angle D$. ∎

Question 3

Sam says: "All rectangles are squares." Give a counter-example to show that Sam is wrong.

▶ Show solution

A rectangle measuring $6$ cm by $2$ cm has four right angles, so it is a rectangle.

Its sides are not all equal, so it is not a square.

One counter-example is enough to disprove the claim. (The correct statement is the reverse: all squares are rectangles.)

Question 4

In triangle $PQR$, $PQ = PR$ and $\angle QPR = 2x$. Prove that $\angle PQR = 90 - x$.

▶ Show solution

$PQ = PR$, so the triangle is isosceles and $\angle PQR = \angle PRQ$ (base angles of an isosceles triangle).

Let each base angle be $y$.

$2x + y + y = 180$ (angle sum of a triangle)

$2y = 180 - 2x$

$y = 90 - x$

Therefore $\angle PQR = 90 - x$. ∎

Question 5

Prove that the diagonals of a rhombus bisect each other at right angles.

▶ Show solution

Let $ABCD$ be a rhombus with diagonals meeting at $O$. A rhombus is a parallelogram, so its diagonals bisect each other: $AO = OC$ and $BO = OD$.

Now consider triangles $AOB$ and $COB$:

$AB = CB$ (all sides of a rhombus are equal)

$AO = CO$ (diagonals bisect each other)

$OB = OB$ (common side)

Therefore $\triangle AOB \cong \triangle COB$ (SSS), so $\angle AOB = \angle COB$.

These two angles lie on the straight line $AOC$, so they add to $180^\circ$. Being equal, each must be $90^\circ$. ∎

Question 6

Two straight lines $AB$ and $CD$ intersect at $X$. Prove that vertically opposite angles are equal, i.e. $\angle AXC = \angle BXD$.

▶ Show solution

$\angle AXC + \angle CXB = 180^\circ$ (angles on the straight line $AB$)

$\angle CXB + \angle BXD = 180^\circ$ (angles on the straight line $CD$)

Therefore $\angle AXC + \angle CXB = \angle CXB + \angle BXD$.

Subtracting $\angle CXB$ from both sides gives $\angle AXC = \angle BXD$. ∎

Question 7

A regular polygon has $n$ sides. Prove that each interior angle is $\dfrac{(n-2) \times 180}{n}$ degrees, and use this to find the interior angle when $n = 8$.

▶ Show solution

The interior angles of any $n$-sided polygon add to $(n-2)\times 180^\circ$.

In a regular polygon all $n$ interior angles are equal, so each one is the total divided by $n$:

$$\text{each interior angle} = \frac{(n-2)\times 180}{n}$$

When $n = 8$: $\dfrac{6 \times 180}{8} = \dfrac{1080}{8} = 135^\circ$. ∎

Question 8

$ABCD$ is a quadrilateral in which $AB \parallel DC$ and $AB = DC$. Prove that $ABCD$ is a parallelogram.

▶ Show solution

Draw the diagonal $AC$ and consider triangles $ABC$ and $CDA$.

$AB = CD$ (given)

$\angle BAC = \angle DCA$ (alternate angles, $AB \parallel DC$)

$AC = CA$ (common side)

Therefore $\triangle ABC \cong \triangle CDA$ (SAS).

Corresponding parts give $BC = DA$, and $\angle BCA = \angle DAC$ — which are alternate angles, so $BC \parallel AD$.

Both pairs of opposite sides are parallel and equal, so $ABCD$ is a parallelogram. ∎

Question 9

In the diagram, $AB \parallel CD$. $\angle BAE = 4x - 10$ and $\angle DCE = x + 50$, and these are alternate angles. Find $x$ and hence both angles.

▶ Show solution

Alternate angles are equal, so:

$4x - 10 = x + 50$

$3x = 60 \Rightarrow x = 20$

$\angle BAE = 4(20) - 10 = 70^\circ$

$\angle DCE = 20 + 50 = 70^\circ$ ✓

Question 10

$ABC$ is an isosceles triangle with $AB = AC$. $D$ is a point on $AB$ and $E$ is a point on $AC$ such that $AD = AE$.

(a) Prove that $\triangle ABE \cong \triangle ACD$. (b) Deduce that $BE = CD$. (c) $BE$ and $CD$ meet at $F$. Prove that triangle $FBC$ is isosceles.

▶ Show solution

(a) $AB = AC$ (given)

$\angle BAE = \angle CAD$ (the same angle, $\angle BAC$, shared by both triangles)

$AE = AD$ (given)

Therefore $\triangle ABE \cong \triangle ACD$ (SAS).

(b) Corresponding sides of congruent triangles are equal, so $BE = CD$. ∎

(c) From the congruence, $\angle ABE = \angle ACD$.

Also $\angle ABC = \angle ACB$ (base angles of the isosceles triangle $ABC$).

Subtracting: $\angle ABC - \angle ABE = \angle ACB - \angle ACD$, that is $\angle FBC = \angle FCB$.

Triangle $FBC$ has two equal angles, so it is isosceles (with $FB = FC$). ∎

Geometric Reasoning & Proof (G6) · GCSE Maths Revision · Created with MathJax