British Flag Theorem

If $ABCD$ is a rectangle and $P$ is any point inside of it, then we have

$PA^2 + PC^2 = PB^2 + PD ^2 .$

This can be proved by using the Pythagorean theorem and projecting point $P$ to the 4 sides of the rectangle.

Drop perpendicular lines from the point $P$ to the sides of the rectangle, meeting sides $AB, BC, CD, AD$ at points $W, X, Y, Z,$ respectively, as shown in the figure. These four points $W, X, Y, Z$ form the vertices of an orthodiagonal quadrilateral. Applying the Pythagorean theorem to the right triangle $AWP$ and observing that $WP = AZ,$ it follows that

$AP^{2} = AW^{2} + WP^{2} = AW^{2} + AZ^{2}.$

By a similar argument, the squared lengths of the distances from $P$ to the other three corners can be calculated as

$\begin{aligned} PC^{2} &= WB^{2} + ZD^{2}\\ BP^{2} &= WB^{2} + AZ^{2}\\ PD^{2} &= ZD^{2} + AW^{2}. \end{aligned}$

Therefore,

$\begin{aligned} AP^{2} + PC^{2} &= \big(AW^{2} + AZ^{2}\big) + \big(WB^{2} + ZD^{2}\big) \\ &= \big(WB^{2} + AZ^{2}\big) + \big(ZD^{2} + AW^{2}\big) \\ &= BP^{2} + PD^{2}.\ _\square \end{aligned}$

(Source: MATHCOUNT 2014)

Point $E$ lies within rectangle $ABCD$. If $AE = 7$, $BE = 5$, and $CE = 8$, what is $DE?$

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Let $DE=x$. By the British flag theorem, we have

$\begin{aligned} 7^2+8^2 &= 5^2+x^2\\ x^2&=88 \\ x&=2\sqrt{22}.\ _\square \end{aligned}$