Difference between revisions of "British Flag Theorem"

m (added picture of flag)
m
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,(0,1));
+
label("$D$",D,(-1,0));
 
dot(D);
 
dot(D);
 
dot(P);
 
dot(P);
label("P",P,(1,1));
+
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 $AP^{2}+PC^{2}=BP^{2}+DP^{2}$. The theorem is called the British flag theorem due to the similarities between the British flag and a diagram of the points (shown below):

British Flag

[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,(-1,0)); dot(D); dot(P); label("$P$",P,NNE); 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 if the point $P$ 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:

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

Therefore:

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

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