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.
- )
Given,
AD ⊥ bisector of BC
∴BD = DC
∠ADB = 90° given
∴ ∠ADC = 180 – ∠ADC [collinear angle]
=180 – 90
= 900
∴ In △ABD and △ADC
i) AD common side
ii) ∠ADB = ∠ADC
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,
In △ABD and △ADC
i) AB = AC [Given]
ii) BD = DC [ ″ ]
iii) AD is common side
∴ △ABD ≅ △ADC by SSS
∴ ∠ABD = ∠ACD [∵ corresponding angles of congruent △s are equal]