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Answer to the 36th question."DON'T SEND STHE SIMILAR QUERIES.KINDLY PERFORM THE STEP BY STEP AND APPROPRIATE ANSWER".

Q.36. In the adjoining figure, AB = AC, D is a point in the interior of $\u25b3$ ABC such that $\angle $ DBC = $\angle $ DCB. Prove that AD bisects $\angle $ BAC of $\u25b3$ ABC.

draw a triangle abc with point d somewhere in the centre. join db, dc and da

B and C are the bases of ur triangle.... u should get a triangle dbc insode the original triangle abc

** GIVEN**: ab = ac, angle dbc = dcb

__ TO PROVE:__AD bisects angle A

** PROOF:** given angle dbc = dcb

Therefore DB = DC [sides opp. equal angles of a triangle]

Consider Triangles ABD, ACD

AB = AC [given]

AD = AD [common]

DB = DC [proven]

Therefore triangle ABD congruent to triangle ACD by SSS cong. rule

By cpct, angle BAD = CAD

Therefore AD bisects angle BAC.

Regards

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