Two circle $\Sigma_1$ and $\Sigma_2$ having centres $C_1$ and $C_2$ intersect at $A$ and $B$. Let $P$ be a point on the segment $AB$ and let $AP\ne{}BP$. The line through $P$ perpendicular to $C_1P$ meets $\Sigma_1$ at $C$ and $D$. The line through $P$ perpendicular to $C_2$ meets $\Sigma_2$ at $E$ and $F$. Prove that $CDEF$ is a rectangle.
Progress: Since $C_1P\perp CD$, $P$ is the midpoint of $CD$ and similarly $P$ is the midpoint of $EF$. Thus $CDEF$ is a parallelogram. How do I finish this problem from here?
I assume that with a line being perpendicular to $C_1$ you mean it perpendicular to $PC_1$.
Hint: Try to justify following equalities and proceed finish from there:
$$PC^2=PC\cdot PD=PA\cdot PB=PE\cdot PF=PE^2$$
So $PC=PE$.