This is a Rectangle Problem from RMO (Regional Mathematics Olympiad) 2015 from Chennai Region.
Problem: Rectangle problem from RMO 2015
Two circles $latex \Sigma_1 &s=2 $ and $latex \Sigma_2 &s=2 $ having centers at $latex C_1 &s=2 $ and $latex C_2 &s=2 $ intersect at A and B. Let P be a point on the segment AB and let $latex AP \neq PB &s=2 $. The line through P perpendicular to $latex C_1 P $ meets $latex \Sigma_1 $ at C and D. The line through P perpendicular to $latex C_2P $ meets $latex \Sigma_2 $ at E and F. prove that C,D, E and F form a rectangle.
Discussion: 
Note that $latex C_1 P \perp CD &s=2$ at P . Since $latex C_1 &s=2$ is center and and CD is chord of the circle $latex \Sigma_1 &s=2$, perpendicular drawn from center to chord bisects the chord. Therefore $latex PC = PD &s=2 $. Similarly since $latex C_2 P \perp EF &s=2 $ at P, hence PE = EF.
Since diagonals bisects each other, clearly $latex CEDF &s=2 $ is a parallelogram.
The power of the point P with respect to the circle $latex \Sigma_2 &s=2 $ is $latex PA \times PB = PE \times PF = PF^2 &s=2 $ since we previously showed PE = PF.
Similarly power of the point P with respect to the circle $latex \Sigma_1 &s=2 $ is $latex PA \times PB = PD \times PC = PD^2 &s=2 $ since we previously showed PC = PD.
Hence we have $latex PA \times PB = PF^2 = PD^2 \Rightarrow PF = PD &s=2 $.
Similarly we can show PE = PC. Thus we have PE = PC = PF = PD.
Now it is trivial to show CEDF is rectangle. (Consider $latex \Delta PEC &s=2 $. Since PE = PC we have $latex \angle PEC = \angle PCE = x &s=2 $ (say). Similarly in $latex \Delta PFC &s=2 $, since PF = PC, we have $latex \angle PFC = \angle PCF = y &s=2 $ (say). Thus $latex 2x + 2y = 180^o $ or $latex \angle PCF + \angle PCE = x + y = 90^o &s=2 $. Similarly one can show that the remaining angle of the quadrilateral is right angle. )
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