Geometric proofs - Year 10 Mathematics

What you will learn

distinguish between demonstration (showing something works for a few cases) and deductive proof (showing it must always work),
write structured proofs using congruent triangle tests (SSS, SAS, AAS, RHS),
apply angle properties of parallel lines and isosceles triangles in proofs,
use proven theorems to solve spatial problems.

Worked example 0 Real-world example: why a diagonal brace works

A rectangular gate $ABCD$ sags without a diagonal brace. A builder adds diagonal $AC$ . Why does this make the gate rigid?

Without the brace, the rectangle can deform into a parallelogram — the angles change while side lengths stay the same.
The diagonal $AC$ creates two triangles: $\triangle ABC$ and $\triangle ACD$ .
Each triangle has three fixed side lengths (the two gate sides plus the brace), so by the SSS condition each triangle is rigid.
Two rigid triangles sharing a common side form a rigid structure.

Key idea: a demonstration would be to push the gate and see it does not move. The proof explains why it cannot move — triangles with fixed sides cannot change shape.

1. Demonstration vs deductive proof

A demonstration (or verification) checks specific cases. For example, measuring the angles of five different triangles and finding that each set sums to $180°$ is a demonstration. It does not prove the result for all triangles.

A deductive proof uses definitions, axioms, and previously proven theorems to establish that a statement must be true in every case. Each step follows logically from the previous ones.

Worked example 1 Demonstration vs proof

Claim: the base angles of an isosceles triangle are equal.

Demonstration: draw three isosceles triangles, measure the base angles, and confirm they are equal.

Proof: let $\triangle ABC$ have $AB = AC$ . Draw the altitude from $A$ to $BC$ , meeting $BC$ at $M$ .

In $\triangle ABM$ and $\triangle ACM$ : $AB = AC$ (given), $AM = AM$ (common side), $\angle AMB = \angle AMC = 90°$ (altitude).
By RHS, $\triangle ABM \cong \triangle ACM$ .
Therefore $\angle ABM = \angle ACM$ (corresponding angles of congruent triangles).

The proof shows the result holds for every isosceles triangle, not just the ones we drew.

2. Proofs involving congruent triangles

Two triangles are congruent if they are identical in shape and size. The four standard tests are:

Test	Condition
SSS	Three pairs of corresponding sides are equal
SAS	Two pairs of sides and the included angle are equal
AAS	Two pairs of angles and a corresponding side are equal
RHS	Right angle, hypotenuse, and one other side are equal

The structure of a congruence proof is:

State which triangles you are comparing.
List the matching parts with reasons.
State the congruence test used.
Draw your conclusion from corresponding parts.

Worked example 2 Proving a quadrilateral property

In quadrilateral $ABCD$ , the diagonals $AC$ and $BD$ bisect each other at point $O$ . Prove that $ABCD$ is a parallelogram.

Since the diagonals bisect each other: $AO = OC$ and $BO = OD$ .
In $\triangle AOB$ and $\triangle COD$ : $AO = OC$ (given), $BO = OD$ (given), $\angle AOB = \angle COD$ (vertically opposite angles).
By SAS, $\triangle AOB \cong \triangle COD$ .
Therefore $AB = CD$ and $\angle OAB = \angle OCD$ (corresponding parts).
Since $\angle OAB = \angle OCD$ , lines $AB$ and $DC$ are parallel (alternate angles with transversal $AC$ ).
Similarly, comparing $\triangle AOD$ and $\triangle COB$ gives $AD = BC$ and $AD \parallel BC$ .
Since both pairs of opposite sides are parallel, $ABCD$ is a parallelogram.

Worked example 3 Proving equal lengths

In $\triangle PQR$ , $S$ is the midpoint of $QR$ , and $PS \perp QR$ . Prove that $PQ = PR$ .

In $\triangle PQS$ and $\triangle PRS$ : $QS = SR$ (S is the midpoint), $PS = PS$ (common), $\angle PSQ = \angle PSR = 90°$ (given).
By SAS, $\triangle PQS \cong \triangle PRS$ .
Therefore $PQ = PR$ (corresponding sides).

3. Angle properties in proofs

Key angle facts used in geometric proofs:

Angles on a straight line sum to $180°$ .
Vertically opposite angles are equal.
Alternate angles (with parallel lines) are equal.
Co-interior angles (with parallel lines) sum to $180°$ .
Base angles of an isosceles triangle are equal.
Angle sum of a triangle is $180°$ .
Exterior angle of a triangle equals the sum of the two non-adjacent interior angles.

Worked example 4 Proof using parallel lines

Lines $AB \parallel CD$ are cut by transversal $EF$ at points $P$ and $Q$ respectively. Prove that the co-interior angles $\angle APQ$ and $\angle PQD$ sum to $180°$ .

$\angle APQ + \angle BPQ = 180°$ (angles on a straight line at $P$ ).
$\angle BPQ = \angle PQD$ (alternate angles, since $AB \parallel CD$ ).
Substituting step 2 into step 1: $\angle APQ + \angle PQD = 180°$ .

Worked example 5 Proof using isosceles triangle properties

Triangle $ABC$ is isosceles with $AB = AC$ . Point $D$ lies on $BC$ such that $AD$ bisects $\angle BAC$ . Prove that $D$ is the midpoint of $BC$ .

In $\triangle ABD$ and $\triangle ACD$ : $AB = AC$ (given), $\angle BAD = \angle CAD$ (AD bisects $\angle BAC$ ), $AD = AD$ (common).
By SAS, $\triangle ABD \cong \triangle ACD$ .
Therefore $BD = CD$ (corresponding sides), so $D$ is the midpoint of $BC$ .

4. Applying theorems to solve spatial problems

Once a theorem is proven, it can be used as a tool in further problems without re-proving it.

Worked example 6 Using congruence to find unknown lengths

In the figure, $ABCD$ is a rectangle. $E$ is the midpoint of $AB$ and $F$ is the midpoint of $CD$ . Prove that $DE = BF$ , and find $DE$ if $AB = 12$ cm and $BC = 5$ cm.

Since $ABCD$ is a rectangle: $AD = BC = 5$ cm, and $AB = DC = 12$ cm.
$E$ is the midpoint of $AB$ , so $AE = 6$ cm and $BE = 6$ cm.
$F$ is the midpoint of $CD$ , so $DF = 6$ cm and $CF = 6$ cm.
In $\triangle ADE$ and $\triangle BCF$ : $AD = BC = 5$ , $AE = BF...$ — wait, we need to set up the correct triangles.

Let us reconsider. In $\triangle ADE$ : $AD = 5$ , $AE = 6$ , $\angle DAE = 90°$ . In $\triangle BCF$ : $BC = 5$ , $CF = 6$ , $\angle BCF = 90°$ . 5. By SAS, $\triangle ADE \cong \triangle BCF$ . 6. Therefore $DE = BF = \sqrt{5^2 + 6^2} = \sqrt{61} \approx 7.81$ cm.

Worked example 7 Angle chasing with proven results

In $\triangle ABC$ , $AB = AC$ and $\angle BAC = 50°$ . Point $D$ lies on $AC$ such that $BD = BC$ . Find $\angle BDC$ .

Since $AB = AC$ , base angles: $\angle ABC = \angle ACB = \dfrac{180° - 50°}{2} = 65°$ .
In $\triangle BDC$ , $BD = BC$ , so $\triangle BDC$ is isosceles with base angles $\angle BDC = \angle BCD$ .
But $\angle BCD = \angle ACB = 65°$ (same angle).
So $\angle BDC = 65°$ and $\angle DBC = 180° - 65° - 65° = 50°$ .

Key congruence tests

SSS

\text{Three pairs of corresponding sides equal} \implies \text{congruent}

SAS

\text{Two sides and the included angle equal} \implies \text{congruent}

AAS

\text{Two angles and a corresponding side equal} \implies \text{congruent}

RHS

\text{Right angle, hypotenuse, and one side equal} \implies \text{congruent}

Practice

Fluency

Tier 1: basic skills

Name the four congruence tests and for each, state what must be equal.
In $\triangle ABC$ and $\triangle DEF$ : $AB = DE$ , $BC = EF$ , $\angle B = \angle E$ . Which congruence test applies?
Triangle $PQR$ has $PQ = PR$ and $\angle QPR = 70°$ . Find $\angle PQR$ and $\angle PRQ$ .
Lines $l_1 \parallel l_2$ are cut by a transversal. One alternate angle is $55°$ . Find the other alternate angle and the co-interior angle on the same side.
In $\triangle XYZ$ , the exterior angle at $Z$ is $130°$ . If $\angle X = 80°$ , find $\angle Y$ .
Explain the difference between a demonstration and a deductive proof using an example.
$ABCD$ is a parallelogram. State two properties of its diagonals that can be proven using congruent triangles.
Triangle $ABC$ has $\angle A = 90°$ , $AB = 6$ , $AC = 8$ , $BC = 10$ . Triangle $DEF$ has $\angle D = 90°$ , $DE = 6$ , $DF = 8$ . Are the triangles congruent? State the test.

Reasoning

Tier 2: structured proofs

In $\triangle ABC$ , $M$ is the midpoint of $BC$ . $AM$ is drawn and $AM = \frac{1}{2}BC$ . Prove that $\angle BAC = 90°$ . (Hint: show $\triangle ABM$ and $\triangle ACM$ are both isosceles.)
$ABCD$ is a kite with $AB = AD$ and $CB = CD$ . Prove that diagonal $AC$ bisects diagonal $BD$ at right angles.
Two circles of equal radius intersect at points $P$ and $Q$ . Prove that the line joining the centres is the perpendicular bisector of $PQ$ .
In $\triangle ABC$ , $D$ is on $BC$ such that $AD \perp BC$ . If $AB = 13$ cm, $AD = 12$ cm, and $DC = 4$ cm, find $BD$ and $AC$ .
Prove that the diagonals of a rhombus bisect each other at right angles.
$\triangle ABC$ has $AB = AC$ . $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = CE$ . Prove that $DE \parallel BC$ .

Reasoning

Tier 3: explain and apply

A student claims that if two triangles have three pairs of equal angles, they must be congruent. Is this correct? Explain your reasoning with an example.
Prove that the angle in a semicircle is $90°$ . (Hint: let $O$ be the centre, and use the isosceles triangle property for $\triangle OAB$ and $\triangle OAC$ where $BC$ is the diameter.)
In quadrilateral $ABCD$ , $AB = CD$ and $AB \parallel CD$ . Prove that $ABCD$ is a parallelogram.
Explain why SSA (two sides and a non-included angle) is not a valid congruence test. Illustrate with two non-congruent triangles that satisfy SSA.

Challenge

Reasoning

Harder reasoning

In $\triangle ABC$ , $\angle A = 60°$ , $AB = AC = 10$ cm. Point $P$ lies inside the triangle such that $PA = PB = PC$ . Find $PA$ . (Hint: $P$ is the circumcentre; use the circumradius formula for a triangle.)
Prove that the medians of a triangle are concurrent. (Hint: let two medians meet at $G$ and show $G$ divides each in the ratio $2:1$ , then prove the third median also passes through $G$ .)
In a cyclic quadrilateral $ABCD$ , prove that opposite angles sum to $180°$ . Use the fact that the angle at the centre is twice the angle at the circumference.
Two triangles $\triangle ABC$ and $\triangle A'B'C'$ have $AB = A'B'$ , $AC = A'C'$ , and $\angle A > \angle A'$ . Prove that $BC > B'C'$ (the hinge theorem). You may use the cosine rule.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Tier 1

SSS: three pairs of corresponding sides equal. SAS: two sides and the included angle equal. AAS: two angles and a corresponding side equal. RHS: right angle, hypotenuse, and one other side equal.
SAS (two sides and the included angle between them).
$\angle PQR = \angle PRQ = \frac{180° - 70°}{2} = 55°$ .
The other alternate angle is $55°$ . The co-interior angle on the same side is $180° - 55° = 125°$ .
Exterior angle $= \angle X + \angle Y$ , so $130° = 80° + \angle Y$ , giving $\angle Y = 50°$ .
A demonstration checks specific cases (e.g. measuring angles in three triangles and finding they sum to $180°$ ). A deductive proof uses logical steps from axioms to show the result holds for all cases. The demonstration could fail for an untested case; the proof cannot.
The diagonals of a parallelogram bisect each other (proven by showing two pairs of congruent triangles using SAS with vertically opposite angles).
Yes, by RHS: right angle, hypotenuse $BC = \sqrt{36 + 64} = 10 =$ hypotenuse of $\triangle DEF$ , and $AB = DE = 6$ .

Tier 2

Since $AM = \frac{1}{2}BC$ and $M$ is the midpoint, $BM = MC = AM$ . So $\triangle ABM$ is isosceles ( $AM = BM$ ) and $\triangle ACM$ is isosceles ( $AM = CM$ ). Let $\angle ABM = \alpha$ and $\angle ACM = \beta$ . In $\triangle ABM$ : $\angle BAM = 180° - 2\alpha$ . In $\triangle ACM$ : $\angle CAM = 180° - 2\beta$ . Also $\alpha + \beta = \angle AMB$ or more directly: $\angle BAC = \angle BAM + \angle CAM = (180° - 2\alpha) + (180° - 2\beta) = 360° - 2(\alpha + \beta)$ . But $\alpha + \beta = \angle ABC + \angle ACB = 180° - \angle BAC$ . So $\angle BAC = 360° - 2(180° - \angle BAC) = 360° - 360° + 2\angle BAC$ , giving $\angle BAC = 2\angle BAC - 0$ … Let us redo cleanly. In $\triangle ABC$ : $\angle A + \alpha + \beta = 180°$ , so $\alpha + \beta = 180° - \angle A$ . In $\triangle ABM$ (isosceles with $AM = BM$ ): $\angle BAM = \alpha$ , so $\angle AMB = 180° - 2\alpha$ . In $\triangle ACM$ (isosceles with $AM = CM$ ): $\angle CAM = \beta$ , so $\angle AMC = 180° - 2\beta$ . But $\angle AMB + \angle AMC = 180°$ (straight line). So $(180° - 2\alpha) + (180° - 2\beta) = 180°$ , giving $360° - 2(\alpha + \beta) = 180°$ , so $\alpha + \beta = 90°$ . Therefore $\angle A = 180° - 90° = 90°$ .
In kite $ABCD$ with $AB = AD$ and $CB = CD$ , let diagonals meet at $O$ . In $\triangle ABC$ and $\triangle ADC$ : $AB = AD$ , $CB = CD$ , $AC$ common. By SSS, $\triangle ABC \cong \triangle ADC$ . So $\angle BAC = \angle DAC$ . Now in $\triangle ABO$ and $\triangle ADO$ : $AB = AD$ , $\angle BAO = \angle DAO$ , $AO$ common. By SAS, $\triangle ABO \cong \triangle ADO$ . So $BO = DO$ (diagonal bisected) and $\angle AOB = \angle AOD = 90°$ (supplementary and equal).
Let the centres be $O_1$ and $O_2$ with equal radius $r$ . Then $O_1P = O_1Q = r$ and $O_2P = O_2Q = r$ . So $O_1$ and $O_2$ are each equidistant from $P$ and $Q$ , meaning both lie on the perpendicular bisector of $PQ$ . The line $O_1O_2$ is therefore the perpendicular bisector of $PQ$ .
$BD = \sqrt{AB^2 - AD^2} = \sqrt{169 - 144} = 5$ cm. $AC = \sqrt{AD^2 + DC^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \approx 12.65$ cm.
In rhombus $ABCD$ , all sides equal. Diagonals $AC$ and $BD$ meet at $O$ . In $\triangle AOB$ and $\triangle COB$ : $AB = CB$ (rhombus), $OB$ common, $AO = OC$ (to prove). Instead: in $\triangle ABD$ and $\triangle CBD$ : $AB = CB$ , $AD = CD$ , $BD$ common. By SSS, $\triangle ABD \cong \triangle CBD$ . So $\angle ABD = \angle CBD$ . Now in $\triangle ABO$ and $\triangle CBO$ : $AB = CB$ , $\angle ABO = \angle CBO$ , $BO$ common. By SAS, congruent. So $AO = CO$ and $\angle AOB = \angle COB = 90°$ .
Let $AD = AE$ (since $BD = CE$ and $AB = AC$ , we have $AD = AB - BD = AC - CE = AE$ ). In $\triangle ADE$ : $AD = AE$ , so $\triangle ADE$ is isosceles with $\angle ADE = \angle AED$ . Also $\angle ADE + \angle AED + \angle A = 180°$ . In $\triangle ABC$ : $\angle ABC = \angle ACB$ (isosceles) and $\angle ABC + \angle ACB + \angle A = 180°$ . So $\angle ADE = \angle ABC$ . Since these are corresponding angles with transversal $AB$ cutting $DE$ and $BC$ , we get $DE \parallel BC$ .

Tier 3

Incorrect. Three equal angle pairs (AAA) guarantee similarity, not congruence. For example, an equilateral triangle with side $3$ cm and an equilateral triangle with side $6$ cm both have all angles $= 60°$ but are clearly not congruent. AAA does not fix the size of the triangle.
Let $BC$ be a diameter with centre $O$ , and $A$ a point on the circle. Then $OA = OB = OC = r$ . In $\triangle OAB$ : $OA = OB$ , so $\angle OAB = \angle OBA = \alpha$ . In $\triangle OAC$ : $OA = OC$ , so $\angle OAC = \angle OCA = \beta$ . In $\triangle ABC$ : $\angle BAC = \alpha + \beta$ and $\angle ABC + \angle BCA + \angle BAC = 180°$ , i.e. $\alpha + \beta + (\alpha + \beta) = 180°$ , so $2(\alpha + \beta) = 180°$ , giving $\angle BAC = 90°$ .
In $\triangle ABC$ and $\triangle CDA$ : $AB = CD$ (given), $AC$ common, $\angle BAC = \angle DCA$ (alternate angles since $AB \parallel CD$ ). By SAS, $\triangle ABC \cong \triangle CDA$ . Therefore $BC = DA$ and $\angle BCA = \angle DAC$ (alternate angles), so $BC \parallel AD$ . Both pairs of opposite sides are parallel, hence $ABCD$ is a parallelogram.
SSA is ambiguous because the given angle is not between the two known sides. For example: $\triangle_1$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ ; $\triangle_2$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ but $\angle B$ can be either acute or obtuse (since $\sin B = \frac{7\sin 30°}{5} = 0.7$ gives $B \approx 44.4°$ or $B \approx 135.6°$ ). Two different triangles satisfy the same SSA conditions.

Challenge

In $\triangle ABC$ with $\angle A = 60°$ and $AB = AC = 10$ , the triangle is isosceles. By the cosine rule: $BC^2 = 100 + 100 - 2(100)\cos 60° = 200 - 100 = 100$ , so $BC = 10$ . The triangle is equilateral. The circumradius $R = \frac{a}{2\sin A} = \frac{10}{2\sin 60°} = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} \approx 5.77$ cm. So $PA = \frac{10\sqrt{3}}{3}$ cm.
Let medians from $A$ and $B$ meet at $G$ . The median from $A$ to midpoint $M$ of $BC$ : by the vector approach, $G = A + \frac{2}{3}(M - A) = \frac{A + 2M}{3} = \frac{A + B + C}{3}$ . Similarly, the median from $B$ to midpoint $N$ of $AC$ gives the same point $G = \frac{A + B + C}{3}$ . The third median from $C$ to midpoint $P$ of $AB$ also passes through $\frac{A + B + C}{3}$ . Since all three medians yield the same intersection point, they are concurrent, and $G$ divides each median in the ratio $2:1$ from vertex.
Let $\angle AOB = 2\alpha$ and $\angle AOD = 2\beta$ where $O$ is the centre (central angles subtended by arcs $AB$ and $AD$ ). Then $\angle ACB = \alpha$ and $\angle ACD = \beta$ (angle at circumference = half central angle). Opposite angle $\angle B =$ angle subtended by arc $ADC$ at circumference. Arc $ADC$ has central angle $2\beta + 2\gamma$ (where $2\gamma$ is the central angle for arc $DC$ ). The key result: $\angle B + \angle D$ corresponds to arcs that together make a full circle ( $360°$ ), so the sum of the half-angles is $180°$ .
By the cosine rule in $\triangle ABC$ : $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos A$ . Similarly $B'C'^2 = A'B'^2 + A'C'^2 - 2 \cdot A'B' \cdot A'C' \cos A'$ . Since $AB = A'B'$ and $AC = A'C'$ , and $\angle A > \angle A'$ , we have $\cos A < \cos A'$ (cosine is decreasing on $[0°, 180°]$ ). Therefore $-2 \cdot AB \cdot AC \cos A > -2 \cdot A'B' \cdot A'C' \cos A'$ , so $BC^2 > B'C'^2$ , hence $BC > B'C'$ .

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