Question

17. In A ABC, AB = AC and the bisectors of
angles B and C intersect at point O. Prove
that :
(i) BO = CO
(ii) AO bisects angle BAC.

note: there is no figure provided​

Answer 1

Step-by-step explanation:

Since the angles opposite to equal sides are equal,

∴AB=AC

⇒∠C=∠B

⇒

2

∠B

=

2

∠C

.

Since BO and CO are bisectors of ∠B and ∠C, we also have

∠ABO=

2

∠B

and ∠ACO=

2

∠C

.

∠ABO=

2

∠B

=

2

∠C

=∠ACO.

Consider △BCO:

∠OBC=∠OCB

⇒BO=CO ....... [Sides opposite to equal angles are equal]

Finally, consider triangles ABO and ACO.

BA=CA ...... (given);

BO=CO ...... (proved);

∠ABO=∠ACO (proved).

Hence, by S.A.S postulate

△ABO≅△ACO

⇒∠BAO=∠CAO⇒AO bisects ∠A.

Answered By Rashmi kushwaha

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Answer 2

$\sf \: To \: prove :- \\ \sf i) \: BO = CO \\ \sf ii) \: AO \: bisects \: \angle \: BAC$

$\sf \: Construction :- \\ \sf \: Draw \: AF\perp \: BC.$

$\sf \: Proof :- \\ \sf \: In \: \triangle \: BOE \: and \: \triangle \: COD, \\ \sf \: BE \: = \: CD \: (given) \\ \sf \: \angle \: BOE = \: \angle \: COD \: (vert. \: oppo. \angle) \\ \sf \angle \: EBO \: = \: \angle \: DCO \: ( \angle \: EBO \: + \angle \: OBF\: = \: \angle \: DCO \: + \angle \: OCF) \\ \\ \sf\therefore \: \triangle \: BOE \: \cong \: \triangle \: COD \: (A.S.A)\\ \sf i)\therefore \: BO \: = \: OC \: (C.P.C.T) \\ \\ \sf \: In \: \triangle \: ABF \: and \: \triangle \: ACF, \\ \sf \: AB \: = \: AC \: (given) \\ \sf \: AF \: = \: AF \: (common) \\ \sf \angle \: ABF \: = \: \angle \: ACF \: ( bisects \: \angle \: A \: and \: \angle \: C) \\ \sf \therefore \: \triangle \: ABF \: \cong\: \triangle \: ACF \: (S.A.S) \\ \sf ii)\therefore \: AF \: bisects \: \angle \: A... \: \: \: (hence \: proved)$