Here it is, ANY help would be appreciated.
If ABCD is a rectangle and P is any point its interior, prove AP˛+PC˛
=PB˛+PD˛
1. Draw a sketch.
2. Those line segments which have the same length are labeled by the same letter.
3. According to the Pythagorean theorem you get:
$\displaystyle \overline{AP}^2+\overline{PC}^2 = r^2 + t^2 +s^2 + u^2$
$\displaystyle {\color{white}\overline{AP}^2+\overline{PC}^2} = \underbrace{r^2 + u^2}_{\overline{PD}^2} + \underbrace{t^2 +s^2}_{\overline{PB}^2}$