# RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2

RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2 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 7.2.

 Board RBSE Class Class 9 Subject Maths Chapter Chapter 7 Chapter Name Congruence and Inequalities of Triangles Exercise Ex 7.2 Number of Questions Solved 11 Category RBSE Solutions

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.2

Question 1.
In figure, AB = AC and ∠B = 58°, then find (RBSESolutions.com) the value of ∠A.

Solution.
∠A = 180° – (58° + 58°) = 180° – 116° = 64°

Question 2.
In figure, AD = BD and ∠C = ∠E. Prove that BC = AE.

Solution.
In ∆AED and ∆BCD
(vertically opposite angles)
∠E = ∠C
∴ ∆AED ≅ ∆BCD
(by AAS congruency property)
=> AE = BC (by c.p.c.t)

Question 3.
If AD be a median of (RBSESolutions.com) an isosceles ABC and ∠A = 120° and AB = AC then find ∠ADB.

Solution.
In ∆ABC
AB = AC (given)
∠B = ∠C = x (say)
x + x + 120° = 180°
=> 2x + 120 = 180°
=> 2x = 60° => x = 30°
∠ADB = 180° – (60° + 30°)

Question 4.
If the bisector of an angle of a triangle (RBSESolutions.com) also bisects the opposite side, show that the triangle is isosceles.
Solution.
Given: In ∆ABC, the bisector of ∠BAC meets BC at D such that AD ⊥ BC

To prove: AB = AC or ∆ABC is an isosceles triangle.
Proof: In ∆’s ABD and ACD
(each is a right angle)
∆ABD ≅ ∆ACD (by AAS congruency property)
=> AB = AC (c.p.c.t)
=> ABC is an isosceles ∆.
Hence proved.

Question 5.
In figure, AB = AC and BE = CD. Prove that AD = AE.

Solution.
In ∆ABD and ∆AEC
BE = DC (given) …(i)
Subtracting DE from (RBSESolutions.com) both side of (i), we get
BE – DE = DC – DE
=> BD = EC
AB = AC (given)
=> ∠B = ∠C
(angle opposite to equal sides are equal)
=> ∆ABD ≅ ∆AEC
=> AD = AE (by c.p.c.t)
Hence proved

Question 6.
Two points E and F of the sides AD and BC respectively of (RBSESolutions.com) a square ABCD such that AF = BE then prove that.
(i) ∠BAF = ∠ABE
(ii) BF = AE

Solution.
∵ ABCD is a square
=> ∠A = ∠B = 90°
Now in ∆AEB and ∆AFB
AF = BE
∠BAE = ∠ABF = 90° (given)
AB = AB (common side)
∴ ∆EAB ≅ ∆FBA
=> ∠FAB = ∠EBA (by c.p.c.t)
and also BF = AE (by c.p.c.t)
Hence proved.

Question 7.
AD and BC are equal perpendiculars to (RBSESolutions.com) a line segment AB (see figure). Show that CD bisects AB.

Solution.
∵ ∠B = ∠A = 90°
(as AD and BC are perpendiculars)
and OB = OA (given)
and ∠BOC = ∠AOD
(vertically opposite angles)
(by AAS congruence rule)
=> OA = OB (by c.p.c.t)
=> CD bisects AB
Hence proved.

Question 8.
In an isosceles triangle (RBSESolutions.com) with AB = AC, the bisectors of ∠B and ∠C meet at O. Produce BO upto M then prove that ∠MOC = ∠ABC.
Solution.

In ∆ABC with AB = AC
∠B = ∠C (given)
In ∆BOC
∠MOC = ∠OBC + ∠OCB

Question 9.
Line l is the bisector of an angle ∠A and ∠B is any point on l. BP and BQ are (RBSESolutions.com) perpendicular from B to the arms of ∠A (see figure). Show that
(i) ∆APB = ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Solution.
(i) In ∆APB and ∆AQB
∠BQA = ∠BPA = 90° (given)
AB = AB (common side)
and ∠QAB = ∠PAB
(∵ l is the bisector of ∠A)
∴ ∆APB ≅ ∆AQB
(by AAS congruence rule)
(ii) ∵ ∆APB ≅ ∆AQB
=> BP = BQ (by c.p.c.t)
i.e. B is equidistant from the arms of ∠A.

Question 10.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC, show that BC = DE.

Solution.
We are given that ∠BAD = ∠EAC
Adding ∠DAC to both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
=> ∠BAC = ∠DAE …(i)
Now in ∆BAC and ∆EAD, we have
∠BAC = ∠DAE [using (i)]
AC = AE (given)
(by SAS congruence rule)
=> BC = DE (by c.p.c.t)
Hence proved.

Question 11.
In right triangle ABC, right angled at C, M is the mid-point (RBSESolutions.com) of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure).

Show that:
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB
(iv) CM = $$\frac { 1 }{ 2 }$$ AB
Solution.
(i) In ∆AMC and ∆BMD
DM = CM (given)
∠3 = ∠4
(vertically opposite angles)
AM = MB
(as M is the mid-point of AB)
=> ∆AMC ≅ ∆BMD
(by SAS congruency rule)

(ii) ∵ ∆AMC ≅ ∆BMD
=> ∠1 = ∠2 (by c.p.c.t)
But they are alternate angles,
so AC || BD
=> ∠ACB + ∠DBC = 180
(sum of the interior angles on (RBSESolutions.com) the same side of a transversal is equal to 180°)
But ∠ACB = 90° (given)
=> ∠DBC = 180° = 90°
(iii) In ∆DBC and ∆ACB
∵ ∠DBC = ∠ACB = 90°
BC = BC (common side)
BD = CA
(∵ ∆AMC ≅ ∆BMD)
∴ ∆DBC ≅ ∆ACB
(by SAS congruency rule)
(iv) ∵ ∆DBC ≅ ∆ACB
=> DC = AB
But M mid-point of DC
=> 2CM = CD
=> 2CM = AB [∵ DC = AB]
=> CM = $$\frac { 1 }{ 2 }$$ AB.
Hence proved.

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