CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 7 TRIANGLES EX – 7.2

Triangles

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that

(i) OB = OC
(ii) AO bisects ∠A

Solution:
(i) in ∆ABC, we have
AB = AC [Given]
∴ ∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q1
⇒ 12∠ABC = 12∠ACB
or ∠OBC = ∠OCB
⇒ OC = OB [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO, we have
AB = AC [Given]
∠OBA = ∠OCA [ ∵12∠B = 12∠C]
OB = OC [Proved above]
∆ABO ≅ ∆ACO [By SAS congruency]
⇒ ∠OAB = ∠OAC [By C.P.C.T.]
⇒ AO bisects ∠A.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q2

Solution:
Since AD is bisector of BC.
∴ BD = CD
Now, in ∆ABD and ∆ACD, we have
AD = DA [Common]
∠ADB = ∠ADC [Each 90°]
BD = CD [Proved above]
∴ ∆ABD ≅ ∆ACD [By SAS congruency]
⇒ AB = AC [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q3

Solution:
∆ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒ ∠BCE = ∠CBF
Now, in ∆BEC and ∆CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴ ∆BEC ≅ ∆CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.]

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q4

Show that
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.

Solution:
(i) In ∆ABE and ∆ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥ AC and CF ⊥ AB]
∠A = ∠A [Common]
BE = CF [Given]
∴ ∆ABE ≅ ∆ACF [By AAS congruency]

(ii) Since, ∆ABE ≅ ∆ACF
∴ AB = AC [By C.P.C.T.]
⇒ ABC is an isosceles triangle.

Question 5.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q5

Solution:
In ∆ABC, we have
AB = AC [ABC is an isosceles triangle]
∴ ∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ∆ are equal]
Again, in ∆BDC, we have
BD = CD [BDC is an isosceles triangle]
∴ ∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒ ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q6

Solution:
AB = AC [Given] …(1)
AB = AD [Given] …(2)
From (1) and (2), we have
AC = AD
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180∘2 = 90°
Thus, ∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.

Solution:
In ∆ABC, we have AB = AC [Given]
∴ Their opposite angles are equal.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q7
⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒ ∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 90∘2 = 45°
Thus, ∠B = 45° and ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.

Solution:
In ∆ABC, we have
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Q8
AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒ ∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴ x + x + x = 180o
⇒ 3x = 180°
⇒ x = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.