Question

In the triangle ABC, AB>AC; BC is the perpendicular bisector of angle BAC intersect at the point D. Prove that BD

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.

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