# Difference between revisions of "British Flag Theorem"

m (added picture of flag) |
m |
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Line 14: | Line 14: | ||

draw(C--P); | draw(C--P); | ||

draw(D--P); | draw(D--P); | ||

− | label("A",A,(-1,0)); | + | label("$A$",A,(-1,0)); |

dot(A); | dot(A); | ||

− | label("B",B,(0,-1)); | + | label("$B$",B,(0,-1)); |

dot(B); | dot(B); | ||

− | label("C",C,(1,0)); | + | label("$C$",C,(1,0)); |

dot(C); | dot(C); | ||

− | label("D",D,(0 | + | label("$D$",D,(-1,0)); |

dot(D); | dot(D); | ||

dot(P); | dot(P); | ||

− | label("P",P, | + | label("$P$",P,NNE); |

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

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

## Revision as of 21:51, 31 March 2014

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):

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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