Exercise 8.3 — Diagonals of Parallelogram

Exercise 8.3 — Diagonal Bisection and Advanced Parallelogram Proofs

Exercise 8.3 from Chapter 8, Quadrilaterals, of Class 9 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) focuses on two important new theorems about diagonals of parallelograms and applies them in nine increasingly challenging problems — including angle calculations, formal congruence proofs, and a culminating proof that a figure formed inside a square is itself a square.

The two new theorems introduced here (diagonals of a parallelogram bisect each other, and its converse) are essential tools that get used repeatedly in the exercise problems and in Class 9 board exams. Understanding when to use the converse — to prove that something is a parallelogram — is equally important.

Key Theorems for This Exercise

Question 1 — Finding Angles When Opposite Angles Are Given as Expressions

In parallelogram ABCD, opposite angles are given as (3x − 2)° and (x + 48)°. Since opposite angles of a parallelogram are equal, set them equal and solve for x.

Question 2 — One Angle is 24° Less Than Twice the Smallest Angle

Here, two adjacent angles of a parallelogram are related by a word description. Let the smaller angle be x; the adjacent angle is "24° less than twice the smaller" → it equals (2x − 24)°. Adjacent angles in a parallelogram are supplementary (sum = 180°).

Question 3 — Prove AF = 2AB (Midpoint of BC, Line Extended)

In parallelogram ABCD, E is the midpoint of BC. DE and AB are both extended to meet at F. The proof uses AAS congruence on triangles formed at E, then applies the parallelogram property CD = AB.

Question 4 — Prove PBCQ is a Parallelogram

P and Q are midpoints of AB and DC respectively in parallelogram ABCD. This question uses the converse theorem: if one pair of opposite sides of a quadrilateral is both equal and parallel, it is a parallelogram.

Question 5 — Isosceles Triangle: ∠DAC = ∠BCA and ABCD is a Parallelogram

ABC is an isosceles triangle (AB = AC). AD bisects the exterior angle QAC at A, and CD ∥ BA. Prove (i) ∠DAC = ∠BCA, and (ii) ABCD is a parallelogram. This proof uses the exterior angle theorem and the property of alternate angles.

Question 6 — Perpendiculars from A and C onto Diagonal BD

In parallelogram ABCD, AP ⊥ BD and CQ ⊥ BD (P and Q are feet of the perpendiculars on diagonal BD). Prove (i) △APB ≅ △CQD, and (ii) AP = CQ.

Question 7 — Two Triangles Joined by Parallel Equal Sides

In triangles ABC and DEF: AB = DE, AB ∥ DE and BC = EF, BC ∥ EF. Corresponding vertices A–D, B–E, C–F are joined. Prove (i) ABED is a parallelogram, (ii) BCFE is a parallelogram, (iii) AC = DF, and (iv) △ABC ≅ △DEF.

Question 8 — Trisection Points of a Diagonal; CQ ∥ AP and AC Bisects PQ

In parallelogram ABCD, P and Q are the points of trisection of diagonal BD (so BP = PQ = QD = ⅓BD). Prove that CQ ∥ AP and that diagonal AC bisects PQ.

Question 9 — Midpoints of a Square's Sides Form Another Square

ABCD is a square. E, F, G, H are midpoints of AB, BC, CD, DA respectively (with AE = BF = CG = DH). Prove that EFGH is a square. This multi-step proof uses SAS congruence to show all sides of EFGH are equal (rhombus), then shows one angle is 90° (making it a square).

Common Mistakes to Avoid

• In Q1 and Q2 — arithmetic errors: After setting up the equation correctly, take care with signs — especially "3x − 2 = x + 48" where moving x gives 2x = 50, not 2x = 46. Always rewrite each step carefully.
• In Q3 — using CD = AB too early: The proof needs CD = BF first (via CPCT), and only then can CD be replaced by AB. Don't skip the CPCT step.
• In Q6 — wrong congruence rule: The proof uses AAS (two angles + non-included side). Writing "by SAS" or "by RHS" would be incorrect here — state the correct rule explicitly.
• In Q9 — stopping at rhombus: Many students prove all sides are equal and then declare EFGH a square. You must prove one angle equals 90° as a separate step.
• Confusing trisection with bisection (Q8): P and Q trisect BD — so BP = PQ = QD = ⅓BD, not ½BD. This means O (the midpoint of BD) lies between P and Q, so OP = OQ, not OP = OQ = OB.

Quick Reference — All Results at a Glance

What This Exercise Prepares You For

Exercise 8.3 is the most proof-heavy exercise in the Quadrilaterals chapter, and the techniques used here — especially AAS congruence with alternate angles (Q3, Q6), the chain of parallelograms approach (Q7), and SAS followed by angle calculation (Q9) — are exactly the methods needed for advanced geometry questions in board exams and competitive entrance tests.

The converse theorems (diagonals bisecting each other → parallelogram; equal and parallel sides → parallelogram) used throughout this exercise reappear frequently in the Triangles chapter's mid-point theorem and in Class 10's similarity and circles chapters. Building fluency with these converses now makes later proof-writing significantly faster.