# Telangana SCERT Class 9 Math Solution Chapter 7 Triangles Exercise 7.2

## Telangana SCERT Solution Class IX (9) Math Chapter 7 Triangles Exercise 7.2

1.) Given.

△ABC isosceles △

AB = AC

∴ ∠ABC = ∠ACB [Opposite angles of corresponding equal sides of △ are equal]

BO bisects ∠ABC

OC bisects ∠ACB

∴ ∠OBC = ∠ABO

∠ACO = ∠OCB.

Now,

∠OBC = ∠OCB [∵∠ABC = ∠ACB and BO and OC are bisects of corresponding angles]

∴ In △ BOC

i) Sides BO = OC [∵ ∠OBC = ∠OCB Opposite sides of equal angle of a △ are equal]

ii) Now,

In △AOB and △AOC

i)AB = AC [Given]

ii) ∠ABO = ∠ACD [∵∠ABC = ∠ACB and OB and OC are bisects]

iii) BO = OC

∴ △AOB ≅ △AOC by SAS

∴ ∠BAO = ∠CAO [∵△AOB ≅△AOC]

Hence, AO bisects ∠A.

1. ) Given,

∴BD = DC

=180 – 90

= 900

iii) BD = DC [∵ AD is bisects of BC]

∴ △ABD ≅ △ADC by SAS

∴ AB = AC [corresponding sides of congruent △ are equal]

∴ △ABC is an isosceles △.

3.) △ABC is an isosceles △

∴ ABC = AC

Now,

In △ABD and △ACE

i) ∠A is common.

ii) ∠ADB = ∠AEC = 90° [∵ BD and CE are altitudes of side AC and AB respectively]

iii) AB = AC [given]

∴ △ABD ≅ △ACE by AAS

∴ BD = CE [∵ Corresponding sides of congruent triangles are equal].

4.) Given,

BD = CE

CE and BD are ⊥ to AB and AC respectively

In △ ABD and △ACE

i) ∠A is common

ii) ∠ADB = ∠AEC = 90° [∵ CE, BD ⊥ To AB, AC respectively]

iii) BD = CE given

i) ∴ △ABD ≅ △ACE by AAS

ii) ∴ AB = AC [∵ corresponding sides of congruent △s are equal]

∴ △ABC is an isosceles △.

5.) Given,

△ ABC and △BCD are isosceles triangle.

∴ AB = AC

BD = DC

Now,

Join point A and D with a line segment.

Now,