# Difference between revisions of "British Flag Theorem"

(Seriously, you guys need to learn asymptote.) |
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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>. | ||

− | + | <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("<math>A</math>",A,(-1,0)); | ||

+ | dot(A); | ||

+ | label("<math>B</math>",B,(0,-1)); | ||

+ | dot(B); | ||

+ | label("<math>C</math>",C,(1,0)); | ||

+ | dot(C); | ||

+ | label("<math>D</math>",D,(0,1)); | ||

+ | dot(D); | ||

+ | dot(P); | ||

+ | label("<math>P</math>",P,(1,1)); | ||

+ | draw((0,85)--(200,85)); | ||

+ | draw((124,0)--(124,150)); | ||

+ | label("<math>w</math>",(124,0),(0,-1)); | ||

+ | label("<math>x</math>",(200,85),(1,0)); | ||

+ | label("<math>y</math>",(124,150),(0,1)); | ||

+ | label("<math>z</math>",(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 | + | The theorem also applies to points outside the rectangle, although the proof is harder to visualize in this case. |

== Proof == | == Proof == | ||

Line 22: | Line 45: | ||

Therefore: | Therefore: | ||

− | * <math>AP^{2} + PC^{2} = Aw^{2} + Az^{2} + wB^{2} + zD^{2} = wB^{2} + Az^{2} + zD^{2} + Aw^{2} = BP^{2} + 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> |

## Revision as of 16:14, 16 October 2007

The **British flag theorem** says that if a point P is chosen inside rectangle ABCD then .

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:

Therefore:

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