Parallel Lines, Triangles & Special Quadrilaterals#
This is one of the most structured topics in Grade 10. Every proof needs a statement and a reason. Markers follow a strict list of acceptable reasons — so you must use the exact correct wording.
Part 1: Parallel Lines & Transversals#
When a transversal crosses two parallel lines, three angle pairs are formed:
| Pattern | Name | Rule | Exam reason |
|---|---|---|---|
| F shape | Corresponding angles | Equal | corresp $\angle$s; $AB \parallel CD$ |
| Z shape | Alternate angles | Equal | alt $\angle$s; $AB \parallel CD$ |
| U shape | Co-interior angles | Sum to $180°$ | co-int $\angle$s; $AB \parallel CD$ |
⚠️ CRITICAL: You MUST state which lines are parallel in the reason. Writing “alt $\angle$s” without naming the parallel lines = zero marks.
The Converse — Proving Lines Are Parallel#
If you can prove that alternate angles are equal (or corresponding angles are equal, or co-interior angles sum to $180°$), then the lines ARE parallel.
| To prove lines parallel | Show that… | Reason |
|---|---|---|
| $AB \parallel CD$ | Alternate angles are equal | converse: alt $\angle$s equal |
| $AB \parallel CD$ | Corresponding angles are equal | converse: corresp $\angle$s equal |
| $AB \parallel CD$ | Co-interior angles sum to $180°$ | converse: co-int $\angle$s suppl |
Other Angle Facts You Need#
| Fact | Rule | Exam reason |
|---|---|---|
| Angles on a straight line | Sum to $180°$ | $\angle$s on a str line |
| Vertically opposite angles | Equal | vert opp $\angle$s |
| Angles around a point | Sum to $360°$ | $\angle$s around a pt |
Worked Example 1 — Finding Angles with Parallel Lines#
$AB \parallel CD$. A transversal crosses both lines. $\hat{A}_1 = 65°$. Find $\hat{C}_1$ (alternate to $\hat{A}_1$) and $\hat{C}_2$ (co-interior with $\hat{A}_1$).
| Statement | Reason |
|---|---|
| $\hat{C}_1 = 65°$ | alt $\angle$s; $AB \parallel CD$ |
| $\hat{C}_2 = 180° - 65° = 115°$ | co-int $\angle$s; $AB \parallel CD$ |
Part 2: Triangle Properties#
Core Properties#
| Property | Rule | Exam reason |
|---|---|---|
| Angle sum | $\hat{A} + \hat{B} + \hat{C} = 180°$ | $\angle$ sum of $\triangle$ |
| Exterior angle | Ext $\angle$ = sum of 2 interior opposite $\angle$s | ext $\angle$ of $\triangle$ |
| Isosceles | 2 equal sides → 2 equal base angles | $\angle$s opp equal sides |
| Converse | 2 equal angles → 2 equal sides | sides opp equal $\angle$s |
| Equilateral | All sides equal → all angles = $60°$ | equilateral $\triangle$ |
Worked Example 2 — Exterior Angle#
In $\triangle PQR$, $\hat{P} = 40°$ and $\hat{Q} = 75°$. Side QR is extended to S. Find $\hat{R}_{\text{ext}}$ (the exterior angle at R).
| Statement | Reason |
|---|---|
| $\hat{R}_{\text{ext}} = \hat{P} + \hat{Q} = 40° + 75° = 115°$ | ext $\angle$ of $\triangle$ |
Check: Interior angle at R = $180° - 40° - 75° = 65°$. And $65° + 115° = 180°$ (angles on a straight line) ✓
Part 3: Triangle Congruence#
Two triangles are congruent if they are identical in shape and size. Once proven congruent, ALL corresponding sides and angles are equal — this is the tool for proving many other things.
The Four Conditions#
| Condition | What you need | Key note |
|---|---|---|
| SSS | 3 pairs of equal sides | |
| SAS | 2 sides + the included angle | The angle MUST be between the two sides |
| AAS | 2 angles + a corresponding side | |
| RHS | Right angle + hypotenuse + one other side | Only for right-angled triangles |
⚠️ AAA does NOT prove congruence — triangles can have the same angles but different sizes (similar, not congruent).
Worked Example 3 — Congruence Proof#
Given: $ABCD$ is a parallelogram. $M$ is the midpoint of $BC$. Prove that $\triangle ABM \equiv \triangle DCM$.
| Statement | Reason |
|---|---|
| $AB = DC$ | opp sides of $\parallel$gram |
| $BM = MC$ | $M$ is midpoint of $BC$ (given) |
| $\hat{B} = \hat{C}$ | opp $\angle$s of $\parallel$gram |
| $\therefore \triangle ABM \equiv \triangle DCM$ | SAS |
Worked Example 4 — Using Congruence to Prove Equal Lengths#
Given: In $\triangle ABC$, $D$ is on $BC$ such that $AD \perp BC$. $AB = AC$. Prove $BD = DC$.
| Statement | Reason |
|---|---|
| $AB = AC$ | Given |
| $AD = AD$ | Common side |
| $A\hat{D}B = A\hat{D}C = 90°$ | $AD \perp BC$ (given) |
| $\triangle ABD \equiv \triangle ACD$ | RHS |
| $\therefore BD = DC$ | Corresponding sides of $\equiv \triangle$s |
Part 4: The Mid-Point Theorem#
The line joining the midpoints of two sides of a triangle is:
- Parallel to the third side
- Half the length of the third side
The Converse#
If a line is drawn through the midpoint of one side of a triangle, parallel to another side, then it bisects the third side.
Worked Example 5#
In $\triangle ABC$, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. $BC = 14$ cm. Find $MN$.
| Statement | Reason |
|---|---|
| $MN \parallel BC$ | Mid-point theorem |
| $MN = \frac{1}{2} \times BC = \frac{1}{2} \times 14 = 7$ cm | Mid-point theorem |
Worked Example 6 — Converse#
In $\triangle PQR$, $M$ is the midpoint of $PQ$ and $MN \parallel QR$ where $N$ is on $PR$. Prove $N$ is the midpoint of $PR$.
| Statement | Reason |
|---|---|
| $M$ is the midpoint of $PQ$ | Given |
| $MN \parallel QR$ | Given |
| $\therefore N$ is the midpoint of $PR$ | Converse of mid-point theorem |
Part 5: Special Quadrilaterals#
The Hierarchy#
A square is a special rectangle, which is a special parallelogram. A rhombus is also a special parallelogram. Understanding the hierarchy helps you know which properties each shape inherits:
$$\text{Parallelogram} \leftarrow \begin{cases} \text{Rectangle} \leftarrow \text{Square} \\ \text{Rhombus} \leftarrow \text{Square} \end{cases}$$A square has ALL the properties of both a rectangle AND a rhombus.
The Complete Property Table#
| Property | Parallelogram | Rectangle | Rhombus | Square | Kite | Trapezium |
|---|---|---|---|---|---|---|
| Opp sides $\parallel$ | ✓ | ✓ | ✓ | ✓ | ✗ | 1 pair |
| Opp sides equal | ✓ | ✓ | ✓ | ✓ | ✗ | ✗ |
| All sides equal | ✗ | ✗ | ✓ | ✓ | ✗ | ✗ |
| Opp angles equal | ✓ | ✓ | ✓ | ✓ | 1 pair | ✗ |
| All angles $90°$ | ✗ | ✓ | ✗ | ✓ | ✗ | ✗ |
| Diag bisect each other | ✓ | ✓ | ✓ | ✓ | ✗ | ✗ |
| Diag equal length | ✗ | ✓ | ✗ | ✓ | ✗ | ✗ |
| Diag $\perp$ | ✗ | ✗ | ✓ | ✓ | ✓ | ✗ |
| Diag bisect angles | ✗ | ✗ | ✓ | ✓ | 1 diag | ✗ |
Proving a Shape is a Parallelogram#
You can prove a quadrilateral is a parallelogram by showing ANY ONE of:
| Method | What to show |
|---|---|
| 1 | Both pairs of opposite sides are parallel |
| 2 | Both pairs of opposite sides are equal |
| 3 | One pair of opposite sides is both parallel and equal |
| 4 | Diagonals bisect each other |
| 5 | Both pairs of opposite angles are equal |
Worked Example 7 — Proving a Parallelogram#
$ABCD$ has $AB = DC$ and $AB \parallel DC$. Prove $ABCD$ is a parallelogram.
| Statement | Reason |
|---|---|
| $AB = DC$ | Given |
| $AB \parallel DC$ | Given |
| $\therefore ABCD$ is a parallelogram | One pair of opp sides both equal and parallel |
Worked Example 8 — Finding Angles in a Parallelogram#
$PQRS$ is a parallelogram. $\hat{P} = 3x + 10°$ and $\hat{R} = 5x - 30°$. Find all angles.
| Statement | Reason |
|---|---|
| $\hat{P} = \hat{R}$ | Opp $\angle$s of $\parallel$gram |
| $3x + 10 = 5x - 30$ | |
| $40 = 2x$ | |
| $x = 20°$ |
$\hat{P} = \hat{R} = 3(20) + 10 = 70°$
$\hat{Q} = \hat{S} = 180° - 70° = 110°$ (co-int $\angle$s; $PQ \parallel SR$)
Check: $70 + 110 + 70 + 110 = 360°$ ✓
Part 6: Areas of Quadrilaterals#
| Shape | Area formula |
|---|---|
| Rectangle | $A = l \times b$ |
| Parallelogram | $A = b \times h$ (height is perpendicular to the base) |
| Triangle | $A = \frac{1}{2} b \times h$ |
| Rhombus | $A = \frac{1}{2} d_1 \times d_2$ (product of diagonals ÷ 2) |
| Kite | $A = \frac{1}{2} d_1 \times d_2$ (same as rhombus) |
| Trapezium | $A = \frac{1}{2}(a + b) \times h$ (average of parallel sides × height) |
Key: The parallelogram height is the perpendicular distance between the parallel sides, NOT the slant side.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Incomplete reasons | “Alt $\angle$s” without naming the parallel lines = 0 marks | Always write: “alt $\angle$s; $AB \parallel CD$” |
| Assuming $90°$ | Never assume an angle is $90°$ unless stated or proven | A shape “looking” square doesn’t make it one |
| Mixing up diagonal properties | Parallelogram: bisect each other (not equal, not $\perp$). Rectangle adds equal. Rhombus adds $\perp$. Square has both | Use the property table |
| SAS — wrong angle | The angle must be between the two sides | If the angle isn’t included, SAS doesn’t apply |
| AAA proves congruence | AAA only proves similarity (same shape), not congruence (same size) | You need at least one pair of equal sides |
| Forgetting mid-point theorem | If you see midpoints on two sides of a triangle → parallel + half length | Scan for midpoints before starting the proof |
| Not checking $360°$ | Angles in any quadrilateral sum to $360°$ | Use this as a check after finding all angles |
💡 Pro Tips for Exams#
1. The $360°$ Check#
The interior angles of any quadrilateral sum to $360°$. After finding all four angles, add them up — if they don’t make $360°$, you have an error.
2. The Proof Template#
Statement | Reason
----------------------------- | -------------------------
[Given fact] | Given
[Derived fact] | [Specific theorem/reason]
[Conclusion] | [Final theorem]3. The “What Can I Use?” Scan#
Before starting a proof, scan the given information for:
- Parallel lines → alternate, corresponding, co-interior angles
- Equal sides → isosceles triangle base angles
- Midpoints → mid-point theorem
- Parallelogram → opposite sides equal, opposite angles equal, diagonals bisect
- Right angle → Pythagoras or RHS congruence
4. The Shape Identification Strategy#
If asked “what type of quadrilateral is $ABCD$?”:
- Check for parallel sides (parallelogram?)
- Check for right angles (rectangle?)
- Check for equal sides (rhombus?)
- Check for both (square?)
📌 Where this leads in Grade 11: Circle Geometry — Euclidean proofs with circles, where cyclic quadrilaterals have their own special angle properties