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.

Answer 1

In ΔPOA and ΔQOA
∠OPA = ∠OQA = 90°
OP = OQ  ...(given)
AO = AO
Therefore, ΔPOA ≅ ΔQOA  ...(SSA criteria)
Hence, ∠PAO = ∠QAO
Thus, OA bisects ∠BAC.