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by **FuadAlAlam** » Mon Mar 01, 2021 1:03 pm

Let the circumcircle of $\triangle PAB$ intersects line $DA$ and $BC$ at points $P_1$ and $P_2$, respectively. Now, $\measuredangle P_1PA = \measuredangle P_1BA = \measuredangle PBA - \measuredangle PBP_1 = \measuredangle PBA - \measuredangle PAP_1 = \measuredangle PBA - \measuredangle PAD = \measuredangle PAD$. Again, $\measuredangle DPP_1 = \measuredangle DPA - \measuredangle P_1PA = \measuredangle DPA - \measuredangle PAD = \measuredangle PBA = \measuredangle PP_1A = \measuredangle PP_1D$. Thus, $DP = P_1D$. Therefore, the internal bisector of $\angle ADP$ is the perpendicular bisector of segment $PP_1$. Similarly, the internal bisector of $\angle PCB$ is the perpendicular bisector of segment $PP_2$. Clearly, the perpendicular bisectors of segments $PP_1, PP_2$ and $AB$ passes through the circumcenter of $\triangle PAB$, as desired.