Solution:
($\Leftarrow$) If $AC = BC$, then the quadrilateral $ABQP$ is cyclic since it is an isosceles trapezoid. If $\angle ACB = 90^\circ$, then we have $\angle ACI = \angle BCI = 45^\circ$, where $I$ is the incenter of $\triangle ABC$. We have also $\angle APC = \angle BQC = 135^\circ$, i.e. $\angle IPC = \angle IQC = 45^\circ$. Therefore $\triangle IPC \sim \triangle ICA$ and $\triangle IQC \sim \triangle ICB$. Hence
$$IP \cdot IA = IC^2 = IQ \cdot IB$$
and therefore the quadrilateral $ABQP$ is cyclic.

($\Rightarrow$) We consider the circumcircle of $\triangle APC$. If it is tangent to $CI$ at the point $C$, then $\angle ACI = \angle IPC = 45^\circ$ and thus $\angle ACB = 90^\circ$. Now let us assume that this circle intersects $CI$ again at some point $R$. Then we have $IP \cdot IA = IR \cdot IC$ and $IP \cdot IA = IQ \cdot IB$. Therefore $IR \cdot IC = IQ \cdot IB$ and the quadrilateral $BCRQ$ is cyclic. Thus
$$\angle BRC = \angle BQC = 135^\circ = \angle APC = \angle ARC.$$
Hence $\triangle ARC \cong \triangle BRC$ and $AC = BC$.