Difference between revisions of "British Flag Theorem"
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| − | The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then <math>AP^{2}+PC^{2}=BP^{2}+DP^{2}</math>. | + | The '''British flag theorem''' says that if a point P is chosen inside [[rectangle]] ABCD then <math>AP^{2}+PC^{2}=BP^{2}+DP^{2}</math>. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below): |
<asy> | <asy> | ||
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P=(124,85); | P=(124,85); | ||
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
| + | draw(A--P); | ||
| + | draw(B--P); | ||
| + | draw(C--P); | ||
| + | draw(D--P); | ||
label("A",A,(-1,0)); | label("A",A,(-1,0)); | ||
dot(A); | dot(A); | ||
Revision as of 23:26, 23 January 2010
The British flag theorem says that if a point P is chosen inside rectangle ABCD then 
. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below):
![[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); draw(A--P); draw(B--P); draw(C--P); draw(D--P); 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/d/e/edeeed9a9014d50477c458b755b5a279e6960089.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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