Question

5.In an isosceles triangle ABC with AB=AC, the bisectors of B and C intersect each
other at O. show that AO bisects A

Answer 1

Answer:

In ∆ABC,

➝ AB = AC

➝ ∠B = ∠C .......[Equation (I)]

In ∆BDC,

➝ AB = AC

➝ ∠B = ∠C ......[From Equation (I)]

➝ ½ ∠B = ½ ∠C

➝ ∠OBC = ∠OCB .......[Equation (II)]

➝ BO = CO ......[Sides of an isosceles ∆]

• Considering ∆ABO and ∆ACO :

➝ AB = AC ......[Given]

➝ AO = AO ......[Common]

➝ BO = OC ......[Equation (III)]

⛬ ∆ABO ≅ ∆ACO

➝ ∠BAD = ∠CAD

⛬ ∠A bisected by OA

Answer 2

Answer: Use the result that angles opposite to equal sides of a triangle are equal and its Converse to show part (i ) & show ΔAOB ≅ ΔAOC by using SAS Congruent rule & then use CPCT for part (ii).

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[fig. is in the attachment]

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.

Step-by-step explanation: hope it will help you

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