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Given ABC is a trianlge in which AB=AC . P is any point in the interior of the triangle such that angle ABP=ACP
In ΔAPB and ΔAPC,
AB=AC[given]
∠ABP=∠ACP [given]
AP=AP[common]
ΔAPB≅ΔAPC[by SAS congruency criterion]
∴∠PAB=∠PAC [corresponding angles of congruent trianlges]
thus, BP=CP
And AP bisects ∠BAC
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