# Difference between revisions of "British Flag Theorem"

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

$[asy] size(300); pair A,B,C,D,P; A=(0,0); B=(200,0); C=(200,150); D=(0,150); P=(124,85); draw(A--B--C--D--cycle); draw(A--P); draw(B--P); draw(C--P); draw(D--P); label("A",A,(-1,0)); dot(A); label("B",B,(0,-1)); dot(B); label("C",C,(1,0)); dot(C); label("D",D,(-1,0)); dot(D); dot(P); label("P",P,NNE); draw((0,85)--(200,85)); draw((124,0)--(124,150)); label("w",(124,0),(0,-1)); label("x",(200,85),(1,0)); label("y",(124,150),(0,1)); label("z",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$

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

Squares of lengths suggest right triangles. We build right triangles by drawing a line through $P$ perpendicular to two sides of the rectangle, as shown below. Both $AXYD$ and $BXYC$ are rectangles. $[asy] pair A,B,C,D,P,X,Y; A = (0,0); B=(1,0); D = (0,0.7); C = B+D; P = (0.3,0.4); X = (0.3,0); Y=(0.3,0.7); draw(A--B--C--D--A--P--C); draw(X--Y); draw(B--P--D); draw(rightanglemark(P,X,A,1.5)); draw(rightanglemark(B,X,P,1.5)); draw(rightanglemark(P,Y,C,1.5)); draw(rightanglemark(D,Y,P,1.5)); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("Y",Y,N); label("X",X,S); label("P",P+(0,0.03),NE);[/asy]$ Applying the Pythagorean Theorem to each of the four right triangles in the diagram, we have \begin{align*}PA^2 &= AX^2+XP^2,\\ PB^2 &= BX^2+XP^2,\\ PC^2 &= CY^2+YP^2,\\ PD^2 &= DY^2+YP^2.\end{align*} So, we have \begin{align*}PA^2+PC^2 &= AX^2+XP^2+CY^2+YP^2,\\ PB^2+PD^2 &= BX^2+XP^2+DY^2+YP^2.\end{align*} From rectangles $AXYD$ and $BXYC$, we have $AX = DY$ and $BX = CY$, so the expressions above for $PA^2 + PC^2$ and $PB^2 + PD^2$ are equal, as desired.