NCERT Solutions class 9 Mathematics 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 = 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

(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]