**NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3**

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

**Ex 7.2 Class 9 Maths Question 1.****
**In an isosceles triangle ABC, with
AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show
that

(i) OB = OC

(ii) AO bisects ∠A

**Solution:**

**(i)**In Î”ABC, AB = AC (Given)

⇒ ∠B = ∠C (∵ Angles opposite to equal sides are equal)

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB

and ∠OBA = ∠OCA …(i) (∵ OB and OC are bisectors of ∠B and ∠C, respectively)

∵ ∠OBC = ∠OCB

⇒ OB = OC …(ii) (∵ Sides opposite to equal angles are equal)

**(ii)** In Î”ABO
and Î”ACO,

AB = AC (Given)

∠OBA = ∠OCA [From eq. (i)]

⇒ OB = OC [From eq. (ii)]

Î”ABO ≅ Î”ACO (By SAS congruence rule)

⇒ ∠BAO = ∠CAO
(By CPCT)

⇒ AO is the bisector of ∠BAC.

**Ex 7.2 Class 9 Maths Question 2.
**In Î”ABC, AD is the perpendicular bisector of BC
(see figure). Show that Î”ABC is an isosceles triangle in which AB = AC.

**Solution:****
Given:** AD is the perpendicular bisector of BC.

**To prove:**Î”ABC is an isosceles triangle, i.e.,

AB = AC

**Proof:**In Î”ADB and Î”ADC,

AD = AD [Common]

BD = DC [Given]

and ∠ADB = ∠ADC [Each= 90°]

Î”ADB ≅ Î”ADC [By SAS congruence rule]

⇒ AB = AC [By CPCT]

So, Î”ABC is an isosceles triangle.

**Ex 7.2 Class 9 Maths 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.

**
Given:** Î”ABC is an isosceles triangle in which AB
= AC, BE ⊥ AC and CF ⊥ AB

**To prove:** BE = CF

**Proof:** In Î”ABE and Î”ACF,

AB = AC [given]

∠AEB = ∠AFC
[each 90 °, BE ⊥ AC and CF ⊥ AB]

∠BAE = ∠CAF [common angle]

∴ Î”ABE ≅ Î”ACF [by AAS
congruence rule]

Then
BE = CF [by CPCT] Hence proved.

**Ex 7.2 Class 9 Maths Question 4.**

ABC is a triangle in which altitudes BE and CF
to sides AC and AB are equal (see figure). Show that

(i) Î”ABE ≅ Î”ACF

(ii) AB = AC, i.e., Î”ABC is an isosceles triangle.

**Solution:****
Given:** Î”ABC in which BE ⊥ AC and CF⊥ AB, such that BE = CF.

**To prove:**(i) Î”ABE ≅ Î”ACF

(ii) AB = AC

**Proof:**(i) In Î”ABC and Î”ACF

∠AEB= ∠AFC [each 90 °]

∠BAE = ∠CAF [common angle]

and
BE = CF [Given]

Î”ABE ≅ Î”ACF [by
AAS congruence rule]

(ii) From part (i), Î”ABE ≅ Î”ACF

AB = AC [by CPCT]

Therefore,
Î”ABC is an isosceles triangle.

**Ex 7.2 Class 9 Maths Question 5.****
**ABC and DBC are two isosceles
triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

**Solution:**

In Î”ABC, we have

AB = AC

⇒ ∠ABC = ∠ACB …(i) [∵ angles opposite to equal sides are equal]

In Î”DBC, we have

BD = CD

⇒ ∠DBC = ∠DCB …(ii) [∵ angles opposite to equal sides are equal]

Adding equations. (i) and (ii), we get

∠ABC + ∠DBC = ∠ACB + ∠DCB

⇒ ∠ABD = ∠ACD Hence, proved.

**Ex 7.2 Class 9 Maths Question 6.****
**Î”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.

**Solution:**

In Î”ABC, AB = AC (Given)

⇒ ∠ACB = ∠ABC ...(i) (∵ Angles opposite to equal sides are equal)

Now, AB = AD (Given)

∴ AC = AD (∵ AB = AC)

Now, in Î”ADC, we have

AD = AC (from above)

⇒ ∠ACD = ∠ADC ...(ii) (∵ Angles opposite to equal sides are equal)

On adding equations. (i) and (ii), we have

∠ACB + ∠ACD = ∠ABC + ∠ADC

⇒ ∠BCD = ∠ABC + ∠BDC (∵ ∠ADC = ∠BDC)

Adding ∠BCD on both sides, we have

∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD

⇒ 2∠BCD = 180° (By angle sum property)

∠BCD = 90°

**Ex 7.2 Class 9 Maths Question 7.
**ABC is a right-angled triangle in which ∠A = 90° and AB
= AC. Find ∠B and ∠C.

**Solution:
**Given, Î”ABC is a right-angled triangle in
which ∠A = 90° and AB = AC.

Then, ∠C = ∠B …(i) [since, angles opposite to equal sides of a triangle are equal]

Now, ∠A + ∠B + ∠C = 180°[since, sum of three angles of a triangle is 180°]

⇒ 90 °+ ∠B + ∠B = 180° [from eq. (i)]

⇒ 2∠B = 90° [∴ ∠B = 45°]

Hence, ∠B = 45° and ∠C = 45°

**Ex 7.2 Class 9 Maths Question 8.**

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

**Solution:**

**Given:** Î”ABC is an equilateral triangle.

**To prove:**∠A = ∠B = ∠C = 60°

**Proof:**Since, all the three angles of an equilateral triangle are equal,

i.e., ∠A = ∠B = ∠C …(i)

∴ ∠A + ∠B + ∠C = 180° [since, sum of three angles of
a triangle is 180°]

⇒ ∠A + ∠A + ∠A = 180° [from eq. (i)]

3∠A = 180°

∠A = 180/3

⇒ ∠A = 60°

Hence, ∠A = ∠B = ∠C = 60°

**Hence proved.**

**Related Links:**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**