British Flag Theorem

Revision as of 04:13, 21 January 2010 by Kamallohia (talk | contribs)

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

[asy] 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,(-1,0)); dot(A); label("B",B,(0,-1)); dot(B); label("C",C,(1,0)); dot(C); label("D",D,(0,1)); dot(D); dot(P); label("P",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("$w$",(124,0),(0,-1)); label("$x$",(200,85),(1,0)); label("$y$",(124,150),(0,1)); label("$z$",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]

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


In Figure 1, by the Pythagorean theorem, we have:

  • $AP^{2} = Aw^{2} + Az^{2}$
  • $PC^{2} = wB^{2} + zD^{2}$
  • $BP^{2} = wB^{2} + Az^{2}$
  • $PD^{2} = zD^{2} + Aw^{2}$


  • $AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} =\nolinebreak BP^{2} +\nolinebreak PD^{2}$

The above result i.e. theorem holds true even if the point P is selected on the boundary of rectangle or even outside the rectangle.

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