Difference between revisions of "British Flag Theorem"
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| − | The theorem also applies | + | The theorem also applies if the point <math>P</math> is selected outside or on the boundary of the rectangle, although the proof is harder to visualize in this case. |
== Proof == | == Proof == | ||
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*<math>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}</math> | *<math>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}</math> | ||
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Revision as of 20:05, 22 January 2010
The British flag theorem says that if a point P is chosen inside rectangle ABCD then 
.
![[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]](http://latex.artofproblemsolving.com/e/9/5/e9500c068d75cb49574a0c6d7b349ae8802dc2b4.png)
The theorem also applies if the point 
is selected outside or on the boundary of the rectangle, although the proof is harder to visualize in this case.
Proof
In Figure 1, by the Pythagorean theorem, we have:
Therefore:
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