# Difference between revisions of "British Flag Theorem"

The British flag theorem says that if a point P is chosen inside rectangle ABCD then $AP^{2}+PC^{2}=BP^{2}+DP^{2}$.

$[asy] size(200); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); label("A",A,(-1,0)); dot(A); label("B",B,(0,-1)); dot(B); label("C",C,(1,0)); dot(C); label("D",D,(0,1)); dot(D); dot(P); label("P",P,(1,1)); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("w",(124,0),(0,-1)); label("x",(200,85),(1,0)); label("y",(124,150),(0,1)); label("z",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$

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:

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

Therefore:

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

The above result i.e. theorem holds true even if the point P is selected on the boundary of rectangle or even outside the rectangle.