RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.3.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Congruence and Inequalities of Triangles |

Exercise |
Ex 7.3 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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)

=> [latex]\frac { 1 }{ 2 }[/latex]BC = [latex]\frac { 1 }{ 2 }[/latex]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 (RBSESolutions.com) 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 (RBSESolutions.com) 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.

We hope the given RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.3 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.3, drop a comment below and we will get back to you at the earliest.

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