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Question
O is any point in the ΔABC such that the perpendicular drawn from O on AB and AC are equal. Prove that OA is the bisector of ∠BAC.
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Solution
In ΔPOA and ΔQOA
∠OPA = ∠OQA = 90°
OP = OQ ...(given)
AO = AO
Therefore, ΔPOA ≅ ΔQOA ...(SSA criteria)
Hence, ∠PAO = ∠QAO
Thus, OA bisects ∠BAC.
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