Grade 10 Mathematics
Grade 10 · Term 2Mathematics

Euclidean Geometry

We revise properties of angles, lines, triangles, and quadrilaterals, then prove congruence, similarity, and the midpoint theorem. Writing formal geometry proofs — stating reasons for every step — is the core skill of this chapter.

4.1 Triangles & Quadrilaterals4.2 Midpoint Theorem & Riders
Week 1

4.1 Triangles & Quadrilaterals

  • Revise basic results: lines, angles, triangles (congruence and similarity)
  • Define and prove properties of the special quadrilaterals: kite, parallelogram, rectangle, rhombus, square, trapezium
  • Prove congruence of triangles (SSS, SAS, AAS, RHS)
🌍

Real-World Connection

Geometry is the language of architecture and engineering. The triangular truss in a bridge is rigid by design — you cannot deform a triangle without changing its side lengths. Every bolt and beam is placed with geometric proof behind it.

Property / Rule

Congruence Conditions (triangles)

Two triangles are congruent (same shape AND size) when any of these hold: SSS (three sides equal), SAS (two sides and included angle equal), AAS (two angles and any corresponding side equal), RHS (right angle, hypotenuse, one other side).

△ABC≡△DEF\triangle ABC \equiv \triangle DEF

Property / Rule

Similarity Conditions (triangles)

Two triangles are similar (same shape, proportional sides) if all three angles are equal (AA is sufficient), or if all three pairs of sides are in the same ratio.

△ABC∥∥△DEF⇒ABDE=BCEF=ACDF\triangle ABC \|\| \triangle DEF \Rightarrow \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}

Properties of special quadrilaterals

  • Parallelogram: opposite sides parallel and equal; opposite angles equal; diagonals bisect each other
  • Rectangle: parallelogram with all right angles; diagonals equal in length
  • Rhombus: parallelogram with all sides equal; diagonals perpendicular and bisect corner angles
  • Square: rectangle AND rhombus; all sides equal, all angles 90°, diagonals perpendicular bisectors
  • Kite: two pairs of adjacent sides equal; one diagonal bisects the other perpendicularly; one axis of symmetry
  • Trapezium: exactly one pair of opposite sides parallel

Area formulas for special quadrilaterals

  • Rectangle: A=l×bA = l \times b (length × breadth)
  • Square: A=s2A = s^2 (side squared)
  • Parallelogram: A=b×hA = b \times h (base × perpendicular height)
  • Rhombus: A=12d1×d2A = \frac{1}{2}d_1 \times d_2 (half product of diagonals) OR A=b×hA = b \times h
  • Trapezium: A=12(a+b)×hA = \frac{1}{2}(a+b) \times h (half sum of parallel sides × perpendicular height)
  • Kite: A=12d1×d2A = \frac{1}{2}d_1 \times d_2 (half product of diagonals)

💡 Tip

When writing geometry proofs, every statement needs a reason in brackets. Acceptable reasons include: given, Pythagoras, exterior angle of a triangle, properties of the specific quadrilateral, etc.

Worked Examples

Worked Example

Prove triangles congruent

Problem

In diagram: AB=DCAB=DC and AB∥DCAB\parallel DC. Prove △ABD≡△CDB\triangle ABD\equiv\triangle CDB.

Worked Example

Prove a quadrilateral is a parallelogram

Problem

ABCD has diagonals that bisect each other at O. Prove ABCD is a parallelogram.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge4 marks
State four properties of a rhombus.
2
L1 · Knowledge1 mark
Name the congruence condition where a right angle, hypotenuse, and one side are equal.
3
L2 · Routine Procedures2 marks
In △ABC\triangle ABC: A^=70°\hat{A}=70°, B^=60°\hat{B}=60°. Find C^\hat{C}.
4
L2 · Routine Procedures3 marks
ABCD is a parallelogram. A^=72°\hat{A}=72°. Find all four angles.
5
L3 · Complex Procedures3 marks
Given △ABC∼△DEF\triangle ABC\sim\triangle DEF with AB=6AB=6, BC=9BC=9, DE=4DE=4. Find EFEF.
6
L3 · Complex Procedures4 marks
The diagonals of parallelogram ABCD intersect at O. Prove △AOB≡△COD\triangle AOB\equiv\triangle COD.
7
L4 · Problem Solving5 marks
ABCD is a square with diagonal AC. Prove △ABC≡△ADC\triangle ABC\equiv\triangle ADC and hence deduce that the diagonals of a square are equal.
8
L4 · Problem Solving5 marks
In a kite ABCDABCD, AB=ADAB=AD and CB=CDCB=CD. Prove that diagonal ACAC is the perpendicular bisector of diagonal BDBD.
Week 2

4.2 Midpoint Theorem & Riders

  • Investigate and prove the midpoint theorem: the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length
  • Apply the midpoint theorem and its converse to solve problems and prove riders
  • Solve problems using properties of parallel lines and triangles
🌍

Real-World Connection

Surveyors use the midpoint theorem when they need to measure across an obstacle. Instead of crossing a river, they find the midpoints of two sides of a triangle formed on their side and measure the shorter connecting segment — half the inaccessible distance.

Property / Rule

Midpoint Theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.

If D=mid(AB) and E=mid(AC), then DE∥BC and DE=12BC\text{If }D=\text{mid}(AB)\text{ and }E=\text{mid}(AC),\text{ then }DE\parallel BC\text{ and }DE=\tfrac{1}{2}BC

Property / Rule

Converse of Midpoint Theorem

If a line is drawn through the midpoint of one side of a triangle parallel to a second side, it bisects the third side.

If DE∥BC and D=mid(AB), then E=mid(AC)\text{If }DE\parallel BC\text{ and }D=\text{mid}(AB),\text{ then }E=\text{mid}(AC)

Property / Rule

Proportionality (similar triangle corollary)

A line drawn parallel to one side of a triangle divides the other two sides proportionally.

ADDB=AEECwhen DE∥BC\frac{AD}{DB}=\frac{AE}{EC}\quad\text{when }DE\parallel BC

💡 Tip

In a rider (proof problem), start from what is given and work step-by-step toward what you need to show. State the theorem being used in brackets at each step — do not skip reasons.

Worked Examples

Worked Example

Apply the midpoint theorem

Problem

In △ABC\triangle ABC: DD is the midpoint of ABAB, EE is the midpoint of ACAC, BC=14BC=14 cm. Find DEDE.

Worked Example

Rider using midpoint theorem

Problem

In △PQR\triangle PQR: SS is the midpoint of PQPQ and TT is the midpoint of PRPR. UU is the midpoint of QRQR. Show that STURSTUR is NOT a parallelogram, but STUSSTUS is. (Actually: prove STQUSTQU is a parallelogram.)
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
In △XYZ\triangle XYZ: AA is the midpoint of XYXY, BB is the midpoint of XZXZ, XY=10XY=10, XZ=8XZ=8 and YZ=12YZ=12. Find ABAB.
2
L1 · Knowledge2 marks
State the converse of the midpoint theorem.
3
L2 · Routine Procedures3 marks
In △ABC\triangle ABC: DD is the midpoint of ABAB, EE is on ACAC with DE∥BCDE\parallel BC. If AC=18AC=18, find ECEC.
4
L2 · Routine Procedures4 marks
In △PQR\triangle PQR: PQ=16PQ=16, QR=20QR=20 and PR=14PR=14. SS and TT are midpoints of PQPQ and PRPR respectively. Find the perimeter of △PST\triangle PST.
5
L3 · Complex Procedures4 marks
In △ABC\triangle ABC: DD and EE are midpoints of ABAB and ACAC. Prove that BCEDBCED is a trapezium.
6
L3 · Complex Procedures3 marks
In △ABC\triangle ABC, DD is on ABAB and EE is on ACAC with ADDB=AEEC=12\dfrac{AD}{DB}=\dfrac{AE}{EC}=\dfrac{1}{2}. What can you conclude about DEDE and BCBC?
7
L4 · Problem Solving6 marks
ABCD is a quadrilateral. P, Q, R, S are midpoints of AB, BC, CD, DA respectively. Prove that PQRS is a parallelogram.
8
L4 · Problem Solving6 marks
In trapezium ABCDABCD, AB∥DCAB\parallel DC. EE is the midpoint of ADAD and FF is the midpoint of BCBC. Prove that EF∥ABEF\parallel AB and EF=12(AB+DC)EF=\tfrac{1}{2}(AB+DC). (Hint: draw diagonal ACAC and let GG be its midpoint.)

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Functions

Euclidean Geometry Grade 10 Maths CAPS Notes & Examples | MathSciBuddy