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British Flag Theorem

From AoPSWiki

The British flag theorem says that if a point P is chosen inside rectangle ABCD then AP^{2}+PC^{2}=BP^{2}+DP^{2}.

size(200);pair A,B,C,D,P;A=(0,0);B=(200,0);C=(200,150);D=(0,150);P=(124,85);draw(A--B--C--D--cycle);label("A",A,(-1...

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

Proof

In Figure 1, by the Pythagorean 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} =\nolinebreak BP^{2} +\nolinebreak PD...

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