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Question
In ΔABC, AB = AC. D is a point in the interior of the triangle such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ΔABC.
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Solution
Since AB = AC
∠ABC = ∠ACB
But ∠DBC = ∠DBC
⇒ ∠ABD = ∠ACD
Now in ΔABD and ΔADC
AB = AC
AD = AD
∠ABD = ∠ACD
Therefore, ΔABD ≅ ΔADC ...(SSA criteria)
Hence, ∠BAD = ∠CAD
Thus, AD bisects ∠BAC.
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