RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3 (Updated for 2024)

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RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3

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Question 1.
In two right triangles, one side has an acute angle of one is equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
Given: In ∆ABC and ∆DEF,
∠B = ∠E = 90°
∠C = ∠F
AB = DE
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 1
To prove: ∆ABC = ∆DEF
Proof: In ∆ABC and ∆DEF,
∠B = ∠E (Each = 90°)
∠C = ∠F (Given)
AB = DE (Given)
∆ABC = ∆DEF (AAS axiom)

Question 2.
If the bisector of the exterior vertical angle of a triangle is parallel to the base. Show that the triangle is isosceles.
Solution:
Given: In ∆ABC, AE is the bisector of vertical exterior ∠A and AE \(\parallel\) BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 2
To prove : ∆ABC is an isosceles
Proof: ∵ AE \(\parallel\) BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle

Question 3.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Given: In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 3
To calculate Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ The sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x= \(\frac { { 180 }^{ \circ } }{ 6 }\) = 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120

Question 4.
Prove that each angle of an equilateral triangle is 60°. [NCERT]
Solution:
Given: ∆ABC is an equilateral triangle
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 4
Proof: In ∆ABC,
AB = AC (Sides of an equilateral triangle)
∴ ∠C = ∠B …(i)
(Angles opposite to equal angles)
Similarly, AB = BC
∴ ∠C = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠A = ∠B = ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\)= 60°

Question 5.
Angles A, B, and C of a triangle ABC are equal to each other. Prove that ∆ABC is equilateral.
Solution:
Given: In ∆ABC, ∠A = ∠B = ∠C
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 5
To prove : ∆ABC is an equilateral
Proof: In ∆ABC,
∴ ∠B = ∠C (Given)
∴ AC = AB …(i) (Sides opposite to equal angles)
Similarly, ∠C = ∠A
∴ BC =AB …(ii)
From (i) and (ii)
AB = BC = CA
Hence ∆ABC is an equilateral triangle

Question 6.
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 90°
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 6
AB =AC (Given)
∴ ∠C = ∠B (Angles opposite to equal sides)
But ∠B + ∠C = 90° (∵ ∠B = 90°)
∴ ∠B = ∠C = \(\frac { { 90 }^{ \circ } }{ 2 }\) = 45°
Hence ∠B = ∠C = 45°

Question 7.
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
Given: In ∆PQR, PQ = PR
S is a point on PQ and PT || QR
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 7
To prove: PS = PT
Proof: ∵ST || QR
∴ ∠S = ∠Q and ∠T = ∠R (Corresponding angles)
But ∠Q = ∠R (∵ PQ = PR)
∴ PS = PT (Sides opposite to equal angles)

Question 8.
In an ∆ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
Given: In ∆ABC, AB = AC the bisectors of ∠B and ∠C intersect at O. M is any point on BO produced.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 8
To prove: ∠MOC = ∠ABC
Proof: In ∆ABC, AB = BC
∴ ∠C = ∠B
∵ OB and OC are the bisectors of ∠B and ∠C
∴ ∠1 =∠2 = \(\frac { 1 }{ 2 }\)∠B
Now in ∠OBC,
Ext. ∠MOC = Interior opposite angles ∠1 + ∠2
= ∠1 + ∠1 = 2∠1 = ∠B
Hence ∠MOC = ∠ABC

Question 9.
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given: In ∆ABC, P is a point on the bisector of ∠B and from P, RPQ || AB is a draw that meets BC in Q
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 – 9
To prove : ∆BPQ is an isosceles
Proof: ∵ BD is the bisector of CB
∴ ∠1 = ∠2
∵ RPQ || AB
∴ ∠1 = ∠3 (Alternate angles)
But ∠1 == ∠2 (Proved)
∴ ∠2 = ∠3
∴ PQ = BQ (sides opposite to equal angles)
∴ ∆BPQ is an isosceles

Question 10.
ABC is a triangle in which ∠B = 2∠C, and D is a point on BC such that AD bisects ∠BAC = 72°.
Solution:
Given: In ∆ABC,
∠B = 2∠C, AD is the bisector of ∠BAC AB = CD
To prove : ∠BAC = 72°
Construction: Draw bisector of ∠B which meets AD at O.

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