Study Materials

# NCERT Solutions for Class 9th Mathematics

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## Chapter 7. Triangles

### Exercise 7.3

EXERCISE- 7.3

1.Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that:

(i) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

Solution:

Given: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.

To prove:

(i) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC

Proof:

In ΔABD and Δ ACD

AB = AC [given]

BD = CD [given]

By SSS Congruence Criterion Rule

Δ ABD Δ ACD

∠ BAP = ∠CAP [CPCT] …

(ii)In ΔABP and Δ ACP

AB = AC [given]

∠ BAP = ∠CAP [proved above]

AP = AP [common]

By SAS Congruence Criterion Rule

Δ ABP Δ ACP

BP = CP [CPCT] … 2

∠APB = ∠APC [CPCT]

(iii)      ∠ BAP = ∠CAP [From eq. 1]

Hence, AP bisects A.

Now, In Δ BDP and Δ CDP

BD = CD [given]

BP = CP [given]

DP = DP [common]

By SSS Congruence Criterion Rule

Δ BDP ≅ Δ CDP

∠ BDP = ∠CDP [CPCT]

AP bisects ∠ D.

(iv) AP stands on B

∠APB + ∠APC = 1800

∠APB +∠APB = 1800[proved above]

∠APB = 1800  /2

∠APB = 900

AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

Solution:

Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.

To prove: (i) AD bisects BC

AB = AC (Given)

By RHS Congruence Criterion Rule

BD = CD (By CPCT)

3. Two sides AB and BC and median AM of one triangle ABC are respectively

equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). Show that:

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Solution:

Given:  Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR.

To prove: (i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Proof: In ∆ABC, AM is the median to BC.

BM = 1/2 BC ... 1

In ∆PQR, PN is the median to QR.

QN = 1/2 QR ... 2

from eq .1 & 2

BM = QN ... 3

Now in ABM and  PQN

AB = PQ (Given)

BM = QN [From equation (3)]

AM = PN [given]

By SSS congruence Criterion rule

∆ABM ≅ ∆PQN

∠B =∠Q [CPCT]

Now in∆ ABC and∆ PQR

AB = PQ [given]

∠B = ∠Q [prove above ]

BC = QR [given]

By SAS congruence Criterion rule

∆ ABC ≅ ∆ PQR

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given:  BE and CF are two equal altitudes of a triangle ABC.

To prove: ABC is a isosceles.

Proof: In ∆BEC and ∆CFB,

BE = CF (Given)

∠BEC = CFB (Each 90°)

BC = CB (Common)

By RHS congruence Criterion rule

∆BEC ≅ ∆CFB

∠BCE = ∠CBF (By CPCT)

AB = AC [Sides opposite to equal angles of a triangle are equal]

Hence, ABC is isosceles.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that:

∠ B = ∠ C.

Solution:

Given:  ABC is an isosceles triangle with AB = AC.

To prove: ∠ B = ∠ C.

Construction: Draw AP ⊥ BC to

Proof :  In ∆APB and ∆APC

∠APB = ∠APC (Each 90º)

AB =AC (Given)

AP = AP (Common)

By RHS Congruence Criterion Rule

∆APB ≅  ∆APC

∠B = ∠C [CPCT]

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