Question

5. Ina AABC, it is given that AB = AC and the bisectors of ZB and C intersect ato. IfM
is a point on BO produced, prove that ZMOC = ZABC.
RC If the line through parallel to AB meets​

Answer 1

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.

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Hope this will help you.....