Exercise 8.2 — Parallelograms

Parallelograms and their properties.

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Exercise 8.2 — Parallelogram Theorems and Proofs

Exercise 8.2 from Chapter 8, Quadrilaterals, of Class 9 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) builds directly on the classification work of Exercise 8.1 by introducing four fundamental theorems about parallelograms and applying them in three formal geometric proofs. Each proof in this exercise is a 4-mark or 5-mark type that frequently appears in board exams.

The key skills practiced here are: stating the correct congruence rule (SSS, RHS, etc.), identifying which properties of the given shape provide the required equal sides/angles, and linking steps logically from "Given" to "Required to Prove."

Diagonal Divides Parallelogram Opposite Sides & Angles Equal Rhombus Diagonal Proof Angle Bisector Theorem

The Four Key Theorems Covered in This Exercise

Before solving the problems, it is essential to understand the four theorems stated in the lesson. These theorems are the "tools" you will use in the proofs — quoting them as reasons within your solutions.

Theorem 1

A diagonal of a parallelogram divides it into two congruent triangles.

In parallelogram ABCD, diagonal BD divides it into △ABD and △CDB, and △ABD ≅ △CDB. This is proved using the ASA congruence rule (alternate interior angles and common side).

Theorem 2

In a parallelogram, opposite sides are equal and opposite angles are equal.

In parallelogram ABCD: AB = DC, AD = BC, ∠A = ∠C, and ∠B = ∠D.

Theorem 3 (Converse)

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

In quadrilateral ABCD, if AB = DC and AD = BC, then ABCD is a parallelogram.

Theorem 4 (Converse)

If each pair of opposite angles of a quadrilateral is equal, then it is a parallelogram.

In quadrilateral ABCD, if ∠A = ∠C and ∠B = ∠D, then ABCD is a parallelogram.

💡 Theorems vs Converses: Theorem 2 says "parallelogram → opposite sides equal." Theorem 3 says "opposite sides equal → parallelogram." These are converses of each other — both are true, but they are separate statements used in different directions of proof.

Question 1 — Prove △AFD ≅ △BEC

Parallelogram ABCD and rectangle ABEF share the common side AB. Points F and E lie on AB (F below A, E below B), so that ABEF forms a rectangle with AF and BE as its widths. The question asks you to prove that the two right-angled triangles △AFD and △BEC are congruent.

Parallelogram ABCD with rectangle ABEF below AB; △AFD and △BEC highlighted

A = top-left of rectangle, B = top-right; D is above A, C above B; F below A, E below B

Question 1

ABCD is a parallelogram and ABEF is a rectangle. Prove that △AFD ≅ △BEC.

Given: ABCD is a parallelogram. ABEF is a rectangle. F and E are on the line below A and B respectively.

In △AFD and △BEC: ∠AFD = ∠BEC = 90° ← each angle of rectangle ABEF is a right angle AD = BC ← opposite sides of parallelogram ABCD are equal AF = BE ← opposite sides of rectangle ABEF are equal By RHS Congruence Rule: ∴ △AFD ≅ △BEC ✓

📌 RHS Rule reminder: RHS (Right angle – Hypotenuse – Side) applies when you have a right angle in both triangles, equal hypotenuses (the slanted sides AD and BC), and one equal leg (AF = BE). The key condition is that the right angle must be confirmed first.

✅ Why both shapes together? The proof needs two shapes because it uses properties from both: AD = BC comes from the parallelogram, while AF = BE and the right angles come from the rectangle. This combined use of two shapes' properties in one proof is a key skill tested in board exams.

Question 2 — Diagonals of a Rhombus Divide It into Four Congruent Triangles

When both diagonals of a rhombus are drawn, they intersect at point O and divide the rhombus into four smaller triangles: △AOB, △BOC, △COD, and △AOD. This question proves all four are congruent — which also explains why the diagonals of a rhombus are perpendicular bisectors of each other.

Rhombus ABCD with diagonals AC and BD intersecting at O, creating 4 triangles

Question 2

Prove that the diagonals of a rhombus divide it into four congruent triangles.

Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.

To Prove: △AOD ≅ △AOB ≅ △COB ≅ △COD

Step 1 — Prove △AOD ≅ △AOB OD = OB ← diagonals of a parallelogram bisect each other; rhombus is a parallelogram AD = AB ← all sides of a rhombus are equal OA = OA ← common side By SSS congruence: △AOD ≅ △AOB ... (1) Step 2 — Prove △AOB ≅ △COB OA = OC ← diagonals bisect each other AB = CB ← all sides of a rhombus are equal OB = OB ← common side By SSS congruence: △AOB ≅ △COB ... (2) Step 3 — Prove △COB ≅ △COD OB = OD ← diagonals bisect each other BC = DC ← all sides of a rhombus are equal OC = OC ← common side By SSS congruence: △COB ≅ △COD ... (3) Conclusion — from (1), (2) and (3): △AOD ≅ △AOB ≅ △COB ≅ △COD ✓ ∴ The diagonals divide the rhombus into four congruent triangles.

📌 SSS Congruence Rule: When all three pairs of sides of two triangles are equal, the triangles are congruent by the Side-Side-Side (SSS) rule. No angle information is needed — just the three equal sides. This is the ideal rule when all sides of the shape are equal (like in a rhombus).

💡 Bonus insight: Since all four triangles are congruent, they have equal areas. This means each triangle has exactly ¼ of the rhombus's area. This idea also confirms that the diagonals of a rhombus bisect each other at right angles (since all four triangles formed are congruent with a shared vertex at O).

Question 3 — Prove ∠COD = ½(∠A + ∠B)

In quadrilateral ABCD, the bisectors of ∠C and ∠D meet at point O (inside the quadrilateral). The question proves that the angle formed at O — that is, ∠COD — equals half the sum of the angles at A and B. This is an elegant result that uses the angle sum property of both a triangle and a quadrilateral.

Quadrilateral ABCD; bisectors of ∠C and ∠D meet at O

Question 3

Bisectors of ∠C and ∠D of quadrilateral ABCD intersect at O. Prove that ∠COD = ½(∠A + ∠B).

Given: In quadrilateral ABCD, OC bisects ∠C and OD bisects ∠D; they meet at O.

To Prove: ∠COD = ½(∠A + ∠B)

Step 1 — Use angle sum in △COD Since OC bisects ∠C: ∠OCD = ½∠C Since OD bisects ∠D: ∠ODC = ½∠D In △COD, angles sum to 180°: ∠OCD + ∠ODC + ∠COD = 180° ½∠C + ½∠D + ∠COD = 180° ∠COD = 180° − ½(∠C + ∠D) ...(1) Step 2 — Use angle sum of the quadrilateral ABCD ∠A + ∠B + ∠C + ∠D = 360° ∠A + ∠B = 360° − (∠C + ∠D) Multiply both sides by ½: ½(∠A + ∠B) = 180° − ½(∠C + ∠D) ...(2) Step 3 — Combine (1) and (2) RHS of (1) = RHS of (2) ∴ ∠COD = ½(∠A + ∠B) ✓

✅ Proof strategy: This is a two-step algebraic proof. First, express ∠COD in terms of ∠C and ∠D using the triangle angle sum. Then, independently express ½(∠A + ∠B) in terms of ∠C and ∠D using the quadrilateral angle sum. When both expressions look the same, the result follows immediately.

⚠️ Common mistake: Students often forget to multiply both sides by ½ in Step 2, or they incorrectly write ∠A + ∠B = 360° − ∠C + ∠D (missing the brackets). The brackets around (∠C + ∠D) are critical — writing it as 360° − ∠C + ∠D gives a completely different (and wrong) expression.

Congruence Rules Used in This Exercise

Exercise 8.2 uses two specific congruence rules — understanding when to use each one is essential for board exam proofs.

Rule	Full Name	Conditions Required	Used In
RHS	Right angle – Hypotenuse – Side	One right angle (90°) in each triangle, equal hypotenuses, and one equal leg.	Q1 (△AFD ≅ △BEC)
SSS	Side – Side – Side	All three pairs of corresponding sides are equal.	Q2 (all three steps for the four rhombus triangles)

💡 How to choose the right rule: If you can see a right angle → try RHS first. If all sides are given as equal (like in a rhombus or square) → use SSS. If you have two angles and one side → use ASA or AAS. Pick the rule that uses the information you actually have, not the one you wish you had.

Common Mistakes to Avoid

Forgetting to state the congruence rule: Every "prove congruent" question requires you to explicitly write the rule (SSS, RHS, ASA, etc.) — not just list the equal sides/angles. Without the rule, the proof is incomplete.
Confusing "diagonals bisect each other" with "diagonals are equal": All parallelograms (including rhombuses) have diagonals that bisect each other. Only rectangles and squares have diagonals that are equal in length.
Bracket errors in Q3: ∠COD = 180° − ½(∠C + ∠D) is very different from 180° − ½∠C + ∠D. Always use brackets when halving a sum.
Wrong justification for AD = BC in Q1: AD = BC here comes from the parallelogram property, not the rectangle. Always state which shape gives each pair of equal sides.
Skipping the "common side" step in Q2: OA = OA, OB = OB, OC = OC are common sides that must be explicitly stated — they are the third equal pair needed for SSS, and omitting them makes the proof invalid.

Quick Reference — Theorems and Proof Outcomes

Theorem / Question	Statement	Method / Result
Theorem 1	Diagonal of parallelogram	Divides into 2 congruent △s (ASA)
Theorem 2	Parallelogram properties	Opposite sides equal; opposite angles equal
Theorem 3	Converse: equal opposite sides	Quadrilateral is a parallelogram
Theorem 4	Converse: equal opposite angles	Quadrilateral is a parallelogram
Q1	△AFD ≅ △BEC (parallelogram + rectangle)	RHS congruence (AD=BC, AF=BE, ∠=90°)
Q2	Rhombus diagonals → 4 congruent △s	SSS (×3 steps): △AOD ≅ △AOB ≅ △COB ≅ △COD
Q3	∠COD = ½(∠A + ∠B)	△ angle sum + quadrilateral angle sum combined

What This Exercise Prepares You For

The proof techniques in Exercise 8.2 — especially the "identify given shape properties → select the correct congruence rule → conclude" pattern — are directly reused in Exercise 8.3 and beyond, where more advanced parallelogram theorems (like the Mid-Point Theorem) are proved. The angle-bisector result from Question 3 also reappears in problems involving angle sums in polygons.

The SSS and RHS congruence techniques practiced here are the same ones used throughout the Triangles chapter, where congruence proofs form the core of the syllabus. Building fluency with proof-writing now makes that chapter significantly easier.

For Telangana and Andhra Pradesh board exams, congruence-based proofs like Q1 and Q2 are classic 4-mark questions. The algebraic angle proof (Q3) tends to appear as a 5-mark problem. Practice writing each step on a separate line with the reason clearly stated in brackets — this maximizes partial credit even if one step is incorrect.

📐 Board Exam Tip (Telangana & AP): For any congruence proof, follow this fixed structure: (1) write "In △___ and △___", (2) list three pairs of equal elements with reasons, (3) state the congruence rule, (4) write the conclusion with ∴. Never skip steps — each numbered line earns a separate mark.