Consider $$\triangle OXP$$ and $$\triangle OYR$$
We know that $$\angle OPX$$ and $$\angle ORY$$ are right angles
So we get
$$\angle OPX=\angle ORY =90^{o}$$
We know that $$OX$$ and $$OY$$ are the radii
$$OX=OY$$
From the figure, we know that the sides of a square are equal
$$OP=OR$$
By RHS congruence criterion
$$\triangle OXP\cong OYR$$
$$PX=RY (c.p.c.t.)$$
We know that $$PQ=QR$$
So we get
$$PQ-PX=QR-RY$$
$$QX=QY$$
Therefore, it is proved that $$QX=QY$$