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Ganita Manjari · Class 9 · Part II · Chapter 12

Ganita Manjari Class 9 Maths Chapter 12 Quadrilaterals Exercise 12.2 Solutions

Four questions on parallelograms: true or false statements, a diagonal that bisects an angle, converses of true properties, and the rectangle made by angle bisectors.

Last updated 6 October 2026

  • 4Questions
  • 0Diagrams
  • Ch 12Quadrilaterals

How to answer every question

  1. Mark what is given Write the equal sides, equal angles or parallel lines on the figure.
  2. Pick a test Use ASA, SAS or SSS for triangles, or the parallelogram tests of Theorems 2 to 5.
  3. Use co-interior or alternate angles They connect parallel sides to angle facts.
  4. Hunt for a counter-example A single example is enough to show that a converse is false.

Rhombus: a parallelogram with equal adjacent sides; its diagonals bisect each other at right angles.

Rectangle: a parallelogram with a right angle; its diagonals are equal.

Converse: the statement with "if" and "then" swapped; it may or may not be true (see Chapter 9 Propositions and their Converses Solutions).

Properties and converses

Chapter 12 Exercise 12.2 Question 1 Solution True or false? (i) A parallelogram with a right angle is a rectangle. (ii) A rhombus with perpendicular diagonals is a square. (iii) If the diagonals of a parallelogram are equal, then it is a rectangle.

Answer: True, False, True
Part (i) · TrueTrue

Adjacent angles of a parallelogram add up to 180°. If one angle is 90°, its neighbours are 90°, and the opposite angles are equal, so all four angles are 90°. That is a rectangle.

Part (ii) · FalseFalse

The diagonals of every rhombus are perpendicular, so this condition says nothing new. A rhombus with side 5 and a 60° angle has perpendicular diagonals but its angles are 60° and 120°, so it is not a square.

What is needed is a right angle (or equal diagonals) as well.

Part (iii) · TrueTrue

Let ABCD be a parallelogram with AC = BD. In ∆ABD and ∆DCA we have AB = DC (opposite sides), AD = DA (common) and BD = CA (given). So ∆ABD ≅ ∆DCA by SSS, and ∠A = ∠D.

Since ∠A + ∠D = 180° for adjacent angles of a parallelogram, each is 90°. By part (i), ABCD is a rectangle.

(i) True. (ii) False. (iii) True.

Chapter 12 Exercise 12.2 Question 2 Solution The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.

Answer: also bisects ∠C; rhombus
AC bisects ∠C

Given ∠BAC = ∠CAD. Since AB ∥ DC with transversal AC, ∠BAC = ∠ACD (alternate angles). Since AD ∥ BC with transversal AC, ∠CAD = ∠ACB (alternate angles).

∠ACD = ∠BAC = ∠CAD = ∠ACB

So ∠ACD = ∠ACB, which means AC bisects ∠C.

ABCD is a rhombus

In ∆ACD, ∠CAD = ∠ACD. Sides opposite equal angles are equal (Statement 2 of Chapter 9 Propositions and their Converses Solutions), so DC = DA.

In a parallelogram, DC = AB and DA = BC. Therefore AB = BC = CD = DA. All four sides are equal, so ABCD is a rhombus.

AC bisects ∠C as well, and ∠CAD = ∠ACD gives DA = DC, so all four sides are equal: ABCD is a rhombus.

Chapter 12 Exercise 12.2 Question 3 Solution Answer Yes or No. If No, add an extra condition so that the answer becomes Yes. (i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus? (ii) If the diagonals bisect each other at right angles, must ABCD be a rhombus? (iii) If the diagonals have equal length, must ABCD be a rectangle?

Answer: Yes, Yes, No
Part (i) · YesYes

AC bisects ∠A and ∠C, so in ∆ABC and ∆ADC: ∠BAC = ∠DAC, ∠BCA = ∠DCA and AC is common. So ∆ABC ≅ ∆ADC by ASA, and AB = AD and CB = CD.

BD bisects ∠B and ∠D, so in the same way ∆BAD ≅ ∆BCD, giving BA = BC and DA = DC.

AB = AD = DC = CB
Part (ii) · YesYes

The diagonals bisect each other, so ABCD is a parallelogram (Theorem 4). Let the diagonals meet at E. In ∆AEB and ∆AED: EB = ED, ∠AEB = ∠AED = 90° and AE is common, so they are congruent by SAS and AB = AD.

A parallelogram with two equal adjacent sides is a rhombus.

Part (iii) · NoNo

Take an isosceles trapezium with A(0, 0), B(4, 0), C(3, 2), D(1, 2). Its diagonals AC and BD both equal √13, but it is not a parallelogram, so it is not a rectangle.

Extra condition: if the diagonals are equal and bisect each other, then ABCD is a parallelogram with equal diagonals, which is a rectangle (Q1 (iii)).

(i) Yes. (ii) Yes. (iii) No; add the condition that the diagonals bisect each other (or that ABCD is a parallelogram).

Chapter 12 Exercise 12.2 Question 4 Solution Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle. Why did we assume AB ≠ BC?

Answer: four right angles
The figure
Parallelogram ABCD with the bisectors of its four angles. They meet at S, P, Q and R, forming a smaller quadrilateral inside.
Q4 · bisectors of the angles of a parallelogram (from the book)
Each angle of PQRS is 90°

Adjacent angles of a parallelogram add up to 180°, so ∠A + ∠D = 180°. The bisectors from A and D meet at S, and in ∆ASD:

∠SAD + ∠SDA = ½(∠A + ∠D) = 90°, so ∠ASD = 180° − 90° = 90°

In the same way the bisectors from D and C meet at P with ∠DPC = 90°, from C and B at Q with ∠CQB = 90°, and from B and A at R with ∠ARB = 90°.

The sides of PQRS lie along these bisectors, so each angle of PQRS (for example ∠PSR) equals one of these right angles. A quadrilateral with four right angles is a rectangle.

Why AB ≠ BC

If AB = BC, the parallelogram is a rhombus. Then the bisector of ∠A is the diagonal AC, which is also the bisector of ∠C, and the bisectors of ∠B and ∠D are the diagonal BD. So all the bisectors pass through the centre, and P, Q, R and S collapse to a single point. There is no rectangle.

Every angle of PQRS is 90°, so it is a rectangle. We need AB ≠ BC because in a rhombus the four bisectors meet at one point.

Answers at a glance

Each statement is proved, or refuted with an example.

QuestionWhat is askedKey ideaAnswer
Q1 True or falseRight angle; equal diagonalsTrue, False, True
Q2 Diagonal bisects ∠AAlternate anglesAC bisects ∠C; rhombus
Q3 ConversesASA, SAS; counter-exampleYes, Yes, No
Q4 Angle bisectors∠A + ∠D = 180°Four right angles; rectangle

Other chapters of Ganita Manjari (Class 9)

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 12, Exercise Set 12.2. The solutions, explanations and diagrams on this page are our own working.

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