Let $O$ be the center of the circumcircle, and $AD$ be the angle bisector of the acute triangle $ABC$. The perpendicular drawn from point $D$ on the line $AO$ intersects the line $AC$ at the point $P$. Prove that $AP = AB$.
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Tags: geometry, equal segments, Circumcenter
Let $O$ be the center of the circumcircle, and $AD$ be the angle bisector of the acute triangle $ABC$. The perpendicular drawn from point $D$ on the line $AO$ intersects the line $AC$ at the point $P$. Prove that $AP = AB$.