Chapter 7 Congruence and Inequalities of Triangles Ex 7.3

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.3

Question 1.
∆ABC and ∆DBC are two isosceles I triangles on the same base BC and vertices A and D are on the same side of BC (see figure). 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 ∆’s ABD and ACD, we have
AB = AC (given)
BD = DC (given)
and AD = AD (common)
∴ ∆ABD ≅ ∆ACD
(by SSS congruency rule)
(ii) In ∆’s ABP and ACP, we have
AB = AC (given)
∠BAP = ∠CAP
and AP = AP (common)
[∵ ∆ABD ≅ ∆ACD => ∠B AD = ∠CAD => ∠BAP = ∠CAP]
∴ ∆ABP ≅ ∆ACP
(by SAS congruency rule)
(iii) We have already proved in (i) that
∆ABD ≅ ∆CAD
=> ∠BAP = ∠CAP
=> AP bisects ∠A i.e. AP is the bisector of ∠A.
In ∆’s BDP and CDP, we have
BD = CD (given)
BP = CP [∵ ∆ABP = ∆ACP]
and DP = DP (common)
∴ ∆BDP ≅ ∆CDP
(by SSS congruency rule)
=> ∠BDP = ∠CDP
=> DP is the bisector of ∠D.
Hence, AP is the bisector of ∠A as well as ∠D.
(iv) In (iii), we have proved that
∆BDP ≅ ∆CDP
=> BP = CP and ∠BPD = ∠CPD = 90°.
∴ ∠BPD and ∠CPD form a linear pair
=> DP is the perpendicular bisector of BC
Hence, 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, we have
AB = AC (given)
AD = AD (common)
∠ADB = ∠ADC = 90° (given)

∴ ∆ABD ≅ ∆ADC
(by RHS congruency rule)
=> BD = DC (by c.p.c.t)
Hence AD bisects BC.
(ii) ∵ ∆ABD ≅ ∆ADC (proved earlier)
=> ∠BAD = ∠DAC (by c.p.c.t)
Hence, AD also bisects ∠A.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to side PQ and QR and median PN of ∆PQR (see figure). Show that
(i) ∆ABM = ∆PQN
(ii) ∆ABC = ∆PQR

Solution.
(i) In ∆’s ABM and PQN
AB = PQ (given)
AM = PN (given)
and BC = QR (given)
=> BC = QR
=> BM = QN
∴ ∆ABM ≅ ∆PQN
(by SSS congruency rule)
(ii) In ∆ABC and ∆PQR
∵ ∆ABM ≅ ∆PQN [proved in (i)]
=> ∠B = ∠Q (by c.p.c.t)
AB = PQ (given)
BC = QR (given)
∴ ∆ABC ≅ ∆PQR
(by SAS congruency rule)

Question 4.
BE and CF are two equal altitudes of ∆ABC. By using RHS congruency rule, prove that ∆ABC is an isosceles triangle.
Solution.

In ∆BFC and ∆CEB
∠BFC = ∠CEB = 90° (given)
hyp. BC = hyp. BC (common)
and altitude CF = altitude BE
=> ∆BFC ≅ ∆CEB
(by RHS congruency rule)
=> ∠B = ∠C
=> ∆ABC is an isosceles triangle.
Hence proved.

Question 5.
∆ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution.
ABC is an isosceles triangle in which AB = AC

Draw AP ⊥ BC
In ∆ABP and ∆ACP
hyp. AB = hyp. AC (given)
AP = AP (common)
and ∠APB = ∠APC = 90° (∵ AP ⊥ BC)
∴ ∆ABP = DACP
(by RHS congruency rule)
=> ∠B = ∠C (by c.p.c.t)
Hence proved.