Practice Maths

L26 — Geometric Proofs

Key Terms

Axiom / given
A statement accepted as true without proof — the starting point of any argument.
Theorem
A statement that has been proved from axioms and other theorems.
Proof
A logical sequence of statements and reasons that establishes a mathematical truth beyond doubt.
Parallel lines
Lines in the same plane that never meet; they create equal alternate angles, equal corresponding angles, and co-interior angles summing to 180°.
Vertically opposite angles
The angles across from each other when two lines intersect — always equal.
Exterior angle
The angle formed outside a triangle when a side is extended — equals the sum of the two non-adjacent interior angles.

What is a geometric proof?

A geometric proof is a logical argument that uses known facts (axioms, theorems, definitions) to establish that a mathematical statement is always true. Each step must be justified.

Key facts used in proofs

FactStatement
Angles on a lineSum = 180°
Angles at a pointSum = 360°
Vertically oppositeEqual
Alternate anglesEqual (parallel lines)
Co-interior anglesSum = 180° (parallel lines)
Corresponding anglesEqual (parallel lines)
Triangle angle sum180°
Exterior angle of ▵= sum of two non-adjacent interior angles
Isosceles ▵Base angles equal; equal sides opposite equal angles
Equilateral ▵All angles 60°; all sides equal

Proof of parallel lines (converse)

If…Then…
Alternate angles are equalLines are parallel
Corresponding angles are equalLines are parallel
Co-interior angles sum to 180°Lines are parallel
a b c d e f g h Corresponding: a=e, b=f, c=g, d=h Alternate: c=f, d=e   Co-interior: c+e=180°
Parallel lines cut by a transversal — angle relationships
Hot Tip: Every step in a proof needs a reason. Use the exact theorem name ("alternate angles, AB ∥ CD") — precise language is what distinguishes a proof from a calculation.

Worked Example 1 — Proving angles equal

AB and CD are straight lines intersecting at O. Prove that ∠AOC = ∠BOD.

Step 1: ∠AOC + ∠BOC = 180° (angles on a straight line AB).

Step 2: ∠BOD + ∠BOC = 180° (angles on a straight line CD).

Step 3: ∴ ∠AOC = 180° − ∠BOC = ∠BOD. (Vertically opposite angles are equal.) □

Worked Example 2 — Exterior angle theorem

In ▵ABC, side BC is extended to D. Prove that ∠ACD = ∠ABC + ∠BAC.

Step 1: ∠BAC + ∠ABC + ∠ACB = 180° (angle sum of triangle).

Step 2: ∠ACB + ∠ACD = 180° (angles on a straight line BCD).

Step 3: ∴ ∠ACD = 180° − ∠ACB = ∠BAC + ∠ABC. □

Worked Example 3 — Proving parallel lines

Two lines are cut by a transversal. The co-interior angles are 112° and 68°. Prove the lines are parallel.

Step 1: 112° + 68° = 180°.

Step 2: When co-interior angles sum to 180°, the lines are parallel (converse of co-interior angles theorem). □

Worked Example 4 — Isosceles triangle proof

In ▵ABC, AB = AC. Prove that ∠ABC = ∠ACB.

Step 1: Let D be the midpoint of BC. Draw AD.

Step 2: In ▵ABD and ▵ACD: AB = AC (given); BD = DC (D is midpoint); AD = AD (common).

Step 3: ∴ ▵ABD ≅ ▵ACD (SSS). ∴ ∠ABD = ∠ACD, i.e. ∠ABC = ∠ACB. □

Worked Example 5 — Two-column proof format

In parallelogram ABCD, prove AB = CD.

StatementReason
∠BAC = ∠DCAAlternate angles (AB ∥ DC)
∠BCA = ∠DACAlternate angles (AD ∥ BC)
AC = CACommon side
▵ABC ≅ ▵CDAAAS
AB = CDCorresponding sides of congruent triangles
See Answers ➔

  1. Angle facts. Fluency

    • (a) Two angles on a straight line are (3x + 10)° and (2x − 5)°. Find x.
    • (b) Three angles at a point are 120°, 95°, and y°. Find y.
    • (c) ∠AOB = 65° (vertically opposite to ∠COD). Find ∠BOC.
    • (d) An exterior angle of a triangle is 118°. One interior non-adjacent angle is 54°. Find the other interior non-adjacent angle.

  2. Parallel lines. Fluency

    • (a) Parallel lines, transversal; one alternate angle = 47°. Find the other alternate angle.
    • (b) Co-interior angles are (4x + 15)° and (3x + 20)°. Find x.
    • (c) Corresponding angles are (5y − 10)° and (3y + 14)°. Find y.
    • (d) Are the lines parallel if alternate angles are 73° and 74°? Justify.

  3. Triangle proofs. Fluency

    • (a) ▵ABC, ∠A = 55°, ∠B = 70°. Find ∠C. Justify using the angle sum theorem.
    • (b) ▵PQR is isosceles with PQ = PR. ∠Q = 38°. Find ∠P. Justify.
    • (c) An exterior angle of an equilateral triangle. Find its size and justify.
    • (d) In ▵ABC, ∠A = ∠B. Prove ▵ABC is isosceles by stating the equal sides.

  4. Proof of congruence. Fluency

    • (a) ▵ABC and ▵DEF: AB = DE = 8, BC = EF = 5, ∠ABC = ∠DEF = 60°. State congruence test and write the congruence statement.
    • (b) ▵PQR and ▵XYZ: ∠P = ∠X = 40°, ∠Q = ∠Y = 75°, PQ = XY = 9. State the test.
    • (c) Two right-angled triangles: hypotenuses 13 cm, one leg each 5 cm. Congruent? State test.
    • (d) ▵ABC and ▵DEF: AB = DE = 7, AC = DF = 10, BC = EF = 6. Congruent? State test.

  5. Parallel lines proof. Understanding

    In the diagram, PQ ∥ RS. Lines AB and CD are transversals. Find angles a, b, c, d.

    P Q R S 65° a b c 50° d
    • (a) Find angle a (alternate to 65°). State the theorem.
    • (b) Find angle b (co-interior with 65°). State the theorem.
    • (c) Find angle c (vertically opposite to angle b). Justify.
    • (d) Find angle d (corresponding to 50°). State the theorem.

  6. Proof — rhombus diagonals bisect at right angles. Understanding

    ABCD is a rhombus (all sides equal). Its diagonals meet at O.

    • (a) Prove ▵AOB ≅ ▵COB using SSS.
    • (b) Hence prove ∠AOB = 90°.
    • (c) Prove that the diagonals bisect each other (AO = CO and BO = DO).
    • (d) What additional property does the rhombus have that a rectangle does not, regarding diagonals?

  7. Two-column proofs. Understanding

    • (a) Prove that vertically opposite angles are equal. (Use angles on a straight line.)
    • (b) Prove the exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
    • (c) In isosceles ▵ABC with AB = AC, prove ∠ABD = ∠ACD where D is the midpoint of BC.
    • (d) Lines l1 and l2 are cut by a transversal. Alternate angles are 52° and 52°. Prove l1 ∥ l2.

  8. Mid-point theorem. Understanding

    In ▵ABC, M and N are the midpoints of AB and AC respectively. The line MN is drawn.

    • (a) Prove ▵AMN ~ ▵ABC.
    • (b) Hence prove MN ∥ BC.
    • (c) Prove MN = ½BC.
    • (d) If BC = 14 cm, find MN. If AM = 5 cm, find AB.

  9. Angle chase in a complex figure. Problem Solving

    In the diagram (not shown), AB ∥ CD. Line EF intersects AB at P and CD at Q. At P, angle APE = 35°. At Q, angle CQF = 48°. Lines PE and QF are extended to meet at R, forming ▵PQR.

    • (a) Find ∠EPQ (angle below AB at P, between AB and PQ).
    • (b) Find ∠FQP (angle above CD at Q, between CD and QP).
    • (c) Use the exterior angle theorem or angle sum to find ∠PRQ.
    • (d) Explain the relationship: ∠PRQ = ∠APE + ∠DQF if AB ∥ CD. Verify with these values.

  10. Proof — Pythagoras’ theorem (geometric approach). Problem Solving

    A square of side (a + b) is constructed. Inside, four congruent right-angled triangles with legs a and b are placed, leaving a central square of side c (the hypotenuse).

    • (a) Write the area of the large square in two ways: as (a + b)², and as the sum of four triangles plus the central square.
    • (b) Show that (a + b)² = 4 × (½ab) + c².
    • (c) Expand and simplify to prove a² + b² = c².
    • (d) Is this a valid proof? What assumptions did we make?