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Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that since the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ and iff $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.

Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ and iff $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.

Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that since the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ or if $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.

Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that since the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ and iff $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.

Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that since the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ or if $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

Assume $\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.

Solution: Construct tangent from $A$ and $C$ and assume it meet at $G$. Similarly construct tangents from $B$ and $D$ to meet at $F$.

Observe that since the perpendicular from $G$ to $AC$ and $F$ to $BD$ intersects at $P,Q$ respectively. Also, $\angle OQB=\angle OQD=\angle OQG=\angle OCG=90^\circ\implies OCGAQ$ is cyclic. Similarly, $OBFDP$ is also cyclic.

Claim: $AG,BD,CG$ are concurrent at $G$.

Proof: Claim is true if $QD$ the angle bisector of $\angle AQC$ intersects the arc $\widehat{AC}$ of $OCGAQ$ at $G$ or if $G$ is the midpoint of the arc. But, it is known that $\triangle ACG$ is isosceles or $G$ is equidistant from $A,C$ since $GA,GC$ are tangents from external point. Clearly, the extension of $DQ$ meets $ABCD$ at $B$. Therefore, $AG,BD,CG$ are concurrent at $G$.

Hence, $ABCD$ is a harmonic quadrilateral, so $BF,DF,CA$ are concurrent at $F$.

$\angle APB=\angle FPB\implies \angle FPB=\angle FDB=\angle FBD=\angle FPD$ since $FB=FD$. Therefore, $\angle APB=\angle APD\implies AC$ bisects $\angle BPD$.