# Need Some Help With A Super Hard Geometry Question

• August 30th 2009, 10:27 AM
Mp5xm8
Need Some Help With A Super Hard Geometry Question
Here it is, ANY help would be appreciated.

If ABCD is a rectangle and P is any point its interior, prove AP²+PC²
=PB²+PD²
• August 30th 2009, 11:11 AM
earboth
Quote:

Originally Posted by Mp5xm8
Here it is, ANY help would be appreciated.

If ABCD is a rectangle and P is any point its interior, prove AP²+PC²
=PB²+PD²

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:

$\overline{AP}^2+\overline{PC}^2 = r^2 + t^2 +s^2 + u^2$

${\color{white}\overline{AP}^2+\overline{PC}^2} = \underbrace{r^2 + u^2}_{\overline{PD}^2} + \underbrace{t^2 +s^2}_{\overline{PB}^2}$