center of square lies on angle bisector (2007 Estonia Open Junior 2.2)

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center of square lies on angle bisector (2007 Estonia Open Junior 2.2)

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#1 h Mar 27, 2020, 10:35 AM

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The center of square $ABCD$ is $K$ . The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$ .

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#2 h Mar 27, 2020, 10:52 AM • 1 Y

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Note that $APKB$ is cyclic.
If $P$ is inside the square, then $\angle BPK=\angle BAK=45^{\circ}=\frac{1}{2}\angle BPx$ where ray $Px$ is opposite to ray $PA$
If $P$ is outside the square, then $\angle BPK=\angle BAK=45^{\circ}=\frac{1}{2}\angle APB$

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