Question

In an isosceles triangle ABC, with AB=AC, the bisectors of angle B and angle C intersect each other at O. Join A to O. Show tha (i) OB=OC (ii) AO bisects angle A

Answer 1

Step-by-step explanation:

GIVEN - ∆ABC is an isosceles ∆ with AB= AC, OB

& OC are the bisectors of angle B and

angle C interest each other at O

i.e, angle OBA= angle OBC

& angle OCA = angle angle OCB

TO PROVE -

1) OB=OC

2) AO bisects angle A

PROOF-

1) in ∆ ABC is an isosceles with AB= AC

angle B= angle C

( Since angles opposite to equal sides are equal )

1/2 angle B= 1/2 angle C

( divide both sides by 2)

angle OBC = angle OCB

& angle OBA= angle OCA...........( 1 )

( angle bisectors )

OB = OC...............( 2 )

( side opposite to the equal angles are equal )

2) in ∆ AOB & ∆AOC

AB= AC....( GIVEN )

angle OBA = angle OCA.....( FROM EQ 1 )

OB = OC.....( FROM EQ 2 )

Therefore ∆AOB =~ ∆AOC

( by SAS congruence rule )

Then,

angle BAO =angle CAO

( by CPCT )

so, AO is the bisector of angle BAC

Answer 2

Step-by-step explanation:

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Solution:-

Given:-

AB = AC and

the bisectors of B and C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

B = C

½ B = ½ C

⇒ OBC = OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects A.

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