British flag theorem

In Euclidean geometry, the British flag theorem says that if a point "P" is chosen inside rectangle "ABCD" then

: $AP^\{2\}+PC^\{2\}=BP^\{2\}+DP^\{2\}.,$

The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case.

Naming

This theorem is named as such because when the involved line segments are drawn, they somewhat resemble a British jack.

Proof

In Figure 1, by Pythagoras theorem, we have:

* $AP^\{2\}\; =\; Aw^\{2\}\; +\; Az^\{2\},$
* $PC^\{2\}\; =\; wB^\{2\}\; +\; zD^\{2\},$
* $BP^\{2\}\; =\; wB^\{2\}\; +\; Az^\{2\},$
* $PD^\{2\}\; =\; zD^\{2\}\; +\; Aw^\{2\},$

Therefore:

: $AP^\{2\}\; +\; PC^\{2\}\; =\; Aw^\{2\}\; +\; Az^\{2\}\; +\; wB^\{2\}\; +\; zD^\{2\}\; =\; wB^\{2\}\; +\; Az^\{2\}\; +\; zD^\{2\}\; +\; Aw^\{2\}\; =\; BP^\{2\}\; +\; PD^\{2\},$

References

* [http://web.mit.edu/hmmt/www/datafiles/solutions/2003/sguts03.pdf Harvard-MIT Mathematics Tournament solutions]
* [http://www.mathlinks.ro/Forum/viewtopic.php?t=112784]