These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3

Question 1.

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

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC

Solution:

(i) In ∆ABD and ∆ACD

AB = AC (∵ ABC is an isosceles ∆)

BD = CD (∵ DBC is an isosceles ∆)

AD = AD (Common)

By S-S-S Congruency condition

∆ABD = ∆ACD

∴ ∠BAD = ∠CAD ……(A) (By C.P.CT)

(ii) In ∆ABP and ∆ACP

AB = AC (∵ ABC is an isoscles ∆)

∠BAP = ∠CAP

AP = AP

By S-A-S Congruency Condition

∠ABP = ∠ACP

Therefore, BP = CP …….(i) (By C.P.C.T.)

(iii) We have

∆ABD = ∆ACD (From (A))

∴ ∠BAD = ∠CAD (By C.P.C.T.)

∴ AP bisects ∠A

Again,

In ∆BDP and ∆GDP

BD = CD (∵ ∆BDC is an isoscles ∆)

DP = DP (Common)

BP = CP (From equation (i))

By S-S-S Congruency Condition

∆BDP ≅ ∆CDP ………(B)

∴ ∠BDP = ∠CDP (By C.P.C.T)

Therefore, AP bisects ∠D …….(iii)

From equation (ii) and (iii) we can say AP bisects ∠A as well as ∠D.

(iv) we have

∆BDP = ∆CDP (from (B))

∴ BP = CP ……(iv) (By C.P.C.T.)

Again, ∠BPD = ∠CPD (By C.P.C.T.)

Now, ∠BPD + ∠CPD = 180

or, ∠BPD + ∠BPD = 180

2∠BPD = 180

∴ ∠BPD = 90 ……(v)

Therefore, From equations (iv) and (v) we can say AP is the perpendicular bisector of BC.

Question 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:

(i) In ∆ABD and ∆ACD

AB = AC (Given)

∠ADB = ∠ADC (each 90°)

AD = AD (Common)

By R.H.S Congruency Condition

∆ABD = ∆ACD ……(i)

∴ BD = CD (By C.P.C.T.)

Therefore, AD bisects BC.

(ii) We have

∆ABD = ∆ACD (from equation (i))

∴ ∠BAD = ∠CAD (By C.P.C.T.)

Therefore, AD bisects ∠A.

Question 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 ∆PQR (see Fig. 7.40). Show that:

(i) ∆ABM ≅ ∆PQN

(ii) ∆ABC ≅ ∆PQR

Solution:

(i) In ∆ABM and ∆PQN

AB = PQ (Given)

BM = QN (Given)

AM = PN (Given)

By S-S-S Congruency Condition,

∆ABM ≅ ∆PQN …..prove (i)

∴ ∠B = ∠Q (By C.P.C.T)

(ii) Now, In ∆ABC and ∆PQR

AB = PQ (Given)

BM = QN (Given)

∴ 2BM = 2QN

or, BC = QR

(Because M and N are respectively the midpoint of side BC and QR)

∠B = ∠Q (Prove above)

By S-A-S Congruency Condition

∆ABC ≅ ∆PQR …….prove (ii)

Question 4.

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

Solution:

In ∆BFC and ∆CEB

CR = BE (Given)

∠BFC = ∠CEB (Each 90°)

BC = BC (Common)

∴ By R-H-S congruency condition,

∆BFC ≅ ∆CEB

∴ ∠B = ∠C (By C.P.C.T)

Therefore, AB = AC (Sides opposite to equal angles of a ∆)

Hence, ∆ABC is isosceles.

Question 5.

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:

In ∆ABP and ∆ACP

AB = AC (Given)

∠APB = ∠APC (Each 90°)

AP = AP (Common)

By R-H-S Congruency Condition,

∆ABP ≅ ∆ACP

or, ∠B = ∠C (By C.P.C.T)