(Uising (11. G3,(03) nd SAS nle).AD ACPCT)EXERCISE 7,2

1.	(Uising (11. G3,(03) nd SAS nle).AD ACPCT)EXERCISE 7,2
Answer» Given: ΔABC is an isosceles∆ with AB = AC, OB & OC are the bisectors of ∠B and ∠C intersect each other at O. i.e, ∠OBA= ∠OBC & ∠OCA= ∠OCB To Prove: i)OB=OC ii)AO bisects ∠A. Proof: (i)In ∆ABC is an isosceles with AB = AC, ∴ ∠B = ∠C [Since , angles opposite to equal sides are equal] ⇒ 1/2∠B = 1/2∠C [Divide both sides by 2] ⇒ ∠OBC = ∠OCB & ∠OBA= ∠OCA.......(1) [Angle bisectors] ⇒ OB = OC .......(2) [Side opposite to the equal angles are equal] (ii)In ΔAOB & ΔAOC, AB = AC (Given) ∠OBA= ∠OCA (from eq1) OB = OC. (from eq 2) Therefore, ΔAOB ≅ ΔAOC ( by SAS congruence rule) Then, ∠BAO = ∠CAO (by CPCT) So, AO is the bisector of ∠BAC.

Answer»

Given:

ΔABC is an isosceles∆ with AB = AC, OB & OC are the bisectors of ∠B and ∠C intersect each other at O. i.e, ∠OBA= ∠OBC & ∠OCA= ∠OCB

To Prove: i)OB=OC ii)AO bisects ∠A.

Proof:

(i)In ∆ABC is an isosceles with AB = AC, ∴ ∠B = ∠C [Since , angles opposite to equal sides are equal]

⇒ 1/2∠B = 1/2∠C [Divide both sides by 2]

⇒ ∠OBC = ∠OCB & ∠OBA= ∠OCA.......(1) [Angle bisectors]

⇒ OB = OC .......(2) [Side opposite to the equal angles are equal]

(ii)In ΔAOB & ΔAOC, AB = AC (Given) ∠OBA= ∠OCA (from eq1) OB = OC. (from eq 2)

Therefore, ΔAOB ≅ ΔAOC ( by SAS congruence rule)

Then, ∠BAO = ∠CAO (by CPCT)

So, AO is the bisector of ∠BAC.

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