British Flag Theorem

In Euclidian Geometry, the British flag theorem states 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$",(0,85),(-1,0)); dot((124,0)); dot((200,85)); dot((124,150)); dot((0,85)); [/asy]$

This is also true when point $P$ is located outside or on the boundary of $ABCD$ , and even when $P$ is located in a Euclidian space where $ABCD$ is embedded.

Proof

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.