$AB$ and $CD$ are two mutually perpendicular chords of a circle of radius $R$. If they intersect at $P$, then show that $AP^2+PB^2+CP^2+PD^2=4R^2.$

Question

$AB$ and $CD$ are two mutually perpendicular chords of a circle of radius $R$. If they intersect at $P$, then show that $AP^2+PB^2+CP^2+PD^2=4R^2.$

This problem was given by a friend and I could not solve it. Hope this problem is not wrong. Thanks for any help.

Answer 1

Note that $PA^2+PC^2=AC^2$ and $PB^2+PD^2=BD^2$. On the other hand, $\angle APC=90^\circ$ implies that the two (small) arcs $\stackrel\frown{AC}$ and $\stackrel{\frown}{BD}$ add up to $180^\circ$. The result follows.

Answer 2

Some hints:

You may assume that the circle is given by $x^2+y^2=R^2$, and that $AB$ is horizontal and $CD$ vertical. This amounts to $$A=(a,v),\quad B=(-a,v),\quad C=(u,c), \quad D=(u,-c)\ ,$$ so that $P=(u,v)$.