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# Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

**Solution:**

Given: AB = PQ, AM = PN, BM = QN

**(i)** In ΔABC, AM is the median to BC.

∴ BM = 1/2 BC

In ΔPQR, PN is the median to QR.

∴ QN = 1/2 QR

It is given that BC = QR

∴ 1/2 BC = 1/2 QR

∴ BM = QN … (1)

In ΔABM and ΔPQN,

AB = PQ (Given)

BM = QN [From equation (1)]

AM = PN (Given)

∴ ΔABM ≅ ΔPQN (Using SSS congruence criterion)

⇒ ∠ABM = ∠PQN (By CPCT)

⇒ ∠ABC = ∠PQR … (2)

**(ii)** In Δ ABC and Δ PQR ,

AB = PQ (Given)

∠ABC = ∠PQR [From Equation (2)]

BC = QR (Given)

∴ ΔABC ≅ ΔPQR (By SAS congruence rule)

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 7

**Video Solution:**

## Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that: (i) Δ ABM ≅ Δ PQN (ii) Δ ABC ≅ Δ PQR

NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.3 Question 3

**Summary:**

If two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR, then ΔABM ≅ ΔPQN by SSS congruence and ΔABC ≅ ΔPQR by SAS congruence.

**☛ Related Questions:**

- ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show thati) ΔABD ≅ ΔACDii) ΔABP ≅ ΔACPiii) AP bisects ∠A as well as ∠Div) AP is the perpendicular bisector of BC.
- AD is an altitude of an isosceles triangle ABC in which AB = AC.Show that i) AD bisects BC ii) AD bisects ∠A.
- BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
- ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

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