# NCERT Solutions for Class 9th Mathematics

## 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]

AD = AD [common]

By SSS Congruence Criterion Rule

Δ ABD ≅ Δ ACD

∠ BAD = ∠CAD [CPCT]

∠ 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 = 180^{0}

∠APB +∠APB = 180^{0}[proved above]

∠APB = 180^{0 }/2

∠APB = 90^{0}

AP is the perpendicular bisector of BC.

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

**(i) AD bisects BC (ii) AD bisects ∠ A.**

**Solution:**

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

**To prove:** (i) AD bisects BC

(ii) AD bisects ∠ A.

** Proof: **In ∆BAD and ∆CAD

∠ ADB = ∠ADC (Each 90º as AD is an altitude)

AB = AC (Given)

AD = AD (Common)

By RHS Congruence Criterion Rule

∆BAD ≅ ∆CAD

BD = CD (By CPCT)

Hence, AD bisects BC.

∠BAD = ∠CAD (By CPCT)

Hence, AD bisects ∠ A

**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]