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Ganita Manjari · Class 9 · Part II · Chapter 12

Ganita Manjari Class 9 Maths Chapter 12 Quadrilaterals Exercise 12.3 Solutions

Five questions on the Midpoint Theorem: the midpoint triangle, a segment bisected by MN, the midpoints of diagonals, and the Varignon parallelogram.

Last updated 6 October 2026

  • 5Questions
  • 0Diagrams
  • Ch 12Quadrilaterals

How to answer every question

  1. Find the midpoints Mark every midpoint and the equal halves.
  2. Use the Midpoint Theorem The segment joining two midpoints is parallel to the third side and half as long.
  3. Use its converse A line through a midpoint parallel to a side bisects the other side.
  4. Finish with a parallelogram Equal and parallel sides, or bisecting diagonals, make a parallelogram.

Midpoint Theorem: PQ ∥ BC and PQ = BC2.

Converse: the line through the midpoint of AB parallel to BC bisects AC.

Varignon parallelogram: the midpoints of the sides of any quadrilateral form a parallelogram.

Midpoint Theorem

Chapter 12 Exercise 12.3 Question 1 Solution P, Q, R are the midpoints of sides AB, AC, BC of ∆ABC. (i) Show that ∆PQR is congruent to ∆QPA and to two other triangles which you should identify. (ii) If ∆ABC is erased, leaving only ∆PQR, can you reconstruct ∆ABC?

Answer: ∆PQR ≅ ∆QPA, ∆RBP, ∆CRQ; yes
The figure
Triangle ABC with the midpoints P of AB, Q of AC and R of BC joined to form the triangle PQR in the middle.
Q1 · the midpoint triangle PQR (from the book)
Part (i) · Using the Midpoint Theorem three times

By the Midpoint Theorem, each segment joining midpoints is half of the third side:

PQ = BC2, QR = AB2, RP = AC2

Also AP = PB = AB2, AQ = QC = AC2, BR = RC = BC2. So QR = AP, RP = AQ, PQ = BR = RC.

TriangleIts three sidesSame as ∆PQR?
∆PQRPQ = BC/2, QR = AB/2, RP = AC/2—
∆QPAQP = BC/2, PA = AB/2, AQ = AC/2yes, by SSS
∆RBPRB = BC/2, BP = AB/2, PR = AC/2yes, by SSS
∆CRQCR = BC/2, RQ = AB/2, QC = AC/2yes, by SSS

So ∆PQR ≅ ∆QPA ≅ ∆RBP ≅ ∆CRQ. The four small triangles cut ∆ABC into four congruent parts.

Part (ii) · Reconstructing ∆ABC

Yes. AB passes through P and is parallel to QR (since PQ and QR are halves of BC and AB with PQ ∥ BC and QR ∥ AB). Likewise BC passes through R parallel to PQ, and CA passes through Q parallel to PR.

  1. Through P draw a line parallel to QR.
  2. Through Q draw a line parallel to RP.
  3. Through R draw a line parallel to PQ.
  4. The three lines form a triangle, which is ∆ABC, with P, Q, R as the midpoints of its sides.

(i) ∆PQR ≅ ∆QPA, ∆PQR ≅ ∆RBP and ∆PQR ≅ ∆CRQ. (ii) Yes: through each vertex of ∆PQR draw a line parallel to the opposite side.

Chapter 12 Exercise 12.3 Question 2 Solution In ∆ABC, M and N are the midpoints of AB and AC. D is any point on BC. Show that MN bisects AD.

Answer: shown
The figure
Triangle ABC with M the midpoint of AB and N the midpoint of AC, and a segment AD from A to a point D on BC crossing MN.
Q2 · MN and the segment AD (from the book)
Proof

Let MN meet AD at X. By the Midpoint Theorem MN ∥ BC, so MX ∥ BD.

Now look at ∆ABD. M is the midpoint of AB and MX is parallel to BD. By the converse of the Midpoint Theorem (Theorem 7), the line through M parallel to BD bisects the third side AD, so X is the midpoint of AD.

MN passes through the midpoint of AD, so MN bisects AD.

Chapter 12 Exercise 12.3 Question 3 Solution In a quadrilateral ABCD, AB ∥ DC and AB ≠ CD. G and H are the midpoints of AC and BD. Prove that GH ∥ AB. (Why did we assume AB ≠ CD?)

Answer: GH ∥ AB
The figure
Quadrilateral ABCD with AB parallel to DC, the diagonals AC and BD, and the midpoints G of AC and H of BD joined.
Q3 · G and H are the midpoints of the diagonals (from the book)
Proof using the midpoint M of AD

Let M be the midpoint of AD.

  • In ∆ACD, M and G are midpoints of AD and AC, so MG ∥ DC and MG = DC2.
  • In ∆ABD, M and H are midpoints of AD and BD, so MH ∥ AB and MH = AB2.

Since AB ∥ DC, both MG and MH are lines through M parallel to AB. There is only one such line, so M, G and H lie on one line, and that line is parallel to AB. Therefore GH ∥ AB. If CD is longer than AB, then GH = MG − MH = CD − AB2.

Why AB ≠ CD

If AB = CD, then AB is equal and parallel to DC, so ABCD is a parallelogram. Its diagonals bisect each other, so G and H are the same point, and there is no segment GH to talk about.

M, G, H lie on a line parallel to AB, so GH ∥ AB. We assume AB ≠ CD so that G and H are different points.

Chapter 12 Exercise 12.3 Question 4 Solution The midpoints of AB, BC, CD, DA of a quadrilateral ABCD are P, Q, R, S. (i) Show that PR and QS bisect each other. (ii) Show that if AC = BD then PR and QS are perpendicular. Is the converse true?

Answer: bisect; perpendicular if AC = BD
Part (i)

By Theorem 9, PQRS is a parallelogram. PR and QS are its diagonals, and the diagonals of a parallelogram bisect each other.

Part (ii)

From the proof of Theorem 9, PQ ∥ AC with PQ = AC2, and QR ∥ BD with QR = BD2. If AC = BD, then PQ = QR.

A parallelogram with two equal adjacent sides is a rhombus, and the diagonals of a rhombus are perpendicular. So PR ⊥ QS.

The converse

Suppose PR ⊥ QS. A parallelogram whose diagonals are perpendicular is a rhombus (Exercise 12.2 Question 3 (ii)), so PQ = QR. Then AC2 = BD2, so AC = BD.

So the converse is true (see Chapter 9 Propositions and their Converses Solutions for how converses are checked).

(i) PR and QS are the diagonals of a parallelogram, so they bisect each other. (ii) If AC = BD then PQRS is a rhombus, so PR ⊥ QS. The converse is true.

Chapter 12 Exercise 12.3 Question 5 Solution PQRS is the Varignon parallelogram of ABCD. (i) Show how to recreate a congruent copy A'B'C'D' of ABCD from PQRS. (ii) Justify that your A'B'C'D' is congruent to ABCD. (iii) Show that if PQRS is a square then AC and BD are perpendicular and equal, and prove the converse.

Answer: place A', then reflect through P, S and R
The figure
Left: quadrilateral ABCD with the midpoints P, Q, R, S of its sides joined. Right: the parallelogram PQRS with a copy A'B'C'D' drawn around it so that P, Q, R, S are again the midpoints of its sides.
Q5 · ABCD with its Varignon parallelogram, and the copy A'B'C'D' (from the book)
Part (i) · Construction

The parallelogram alone does not decide the shape: many quadrilaterals have the same Varignon parallelogram. So the position of A' must be taken from ABCD, in the same way that A sits relative to P and S.

  1. Place A' in the position that A has relative to PQRS (the copy of ABCD is a translation of ABCD, and PQRS is copied by the same translation).
  2. Reflect A' in P to get B', so that P is the midpoint of A'B'.
  3. Reflect A' in S to get D', so that S is the midpoint of A'D'.
  4. Reflect D' in R to get C'.

Another way: turn the whole quadrilateral ABCD through 180° about the centre O of PQRS. The turned copy has the same Varignon parallelogram PQRS (the parallelogram looks the same after a half-turn), only the labels of P, Q, R, S shift by two places.

Part (ii) · Why it is congruent, and why S is on A'D'

By construction, S is the midpoint of A'D' (so S, A', D' are collinear), P is the midpoint of A'B' and R is the midpoint of C'D'. For Q, add the equal vectors: B' + C' = (2P − A') + (2R − D') = 2P + 2R − 2S, and since PQRS is a parallelogram, P + R = Q + S. So B' + C' = 2Q, and Q is the midpoint of B'C'.

Since the copy is a translation of ABCD (or a half-turn of it), it is congruent to ABCD. In the translated figure, ∆S'D'R' is just a translated ∆SDR, so ∆SDR ≅ ∆S'D'R' as the hint says.

Part (iii) · PQRS is a square

PQ ∥ AC with PQ = AC2, and QR ∥ BD with QR = BD2.

  • PQRS is a square exactly when PQ = QR and ∠PQR = 90°.
  • PQ = QR means AC2 = BD2, that is AC = BD.
  • ∠PQR = 90° means PQ ⊥ QR, and since PQ ∥ AC and QR ∥ BD, that means AC ⊥ BD.

Each step works in both directions, so the converse is also true: if AC and BD are equal and perpendicular then PQRS is a parallelogram with equal adjacent sides and a right angle, which is a square.

(i), (ii) Place A', reflect it in P and S, then reflect D' in R; the copy is congruent to ABCD. (iii) PQRS is a square ⟺ AC = BD and AC ⊥ BD.

Answers at a glance

Every answer comes from the Midpoint Theorem or its converse.

QuestionWhat is askedKey ideaAnswer
Q1 Midpoint triangleMidpoint Theorem; SSS∆QPA, ∆RBP, ∆CRQ; can rebuild ABC
Q2 MN bisects ADConverse of Midpoint TheoremShown
Q3 GH ∥ ABUse the midpoint M of ADShown; AB ≠ CD so G ≠ H
Q4 Diagonals of PQRSParallelogram and rhombusBisect; perpendicular if AC = BD; converse true
Q5 Recreate ABCDReflect in P, S, RConstruction; square ⟺ AC = BD and AC ⊥ BD

Other chapters of Ganita Manjari (Class 9)

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 12, Exercise Set 12.3. The solutions, explanations and diagrams on this page are our own working.

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