Two overlapping squares

Question

$ABCD$ is a square. $BEFG$ is another square drawn with the common vertex $B$ such that $E,\ F$ fall inside the square $ABCD$. Then prove that $DF^2=2\cdot AE^2$.

Answer 1

Let $ABCD$ be the unit square with $A =(0,0)$, and let $E = (x,y)$. Then $\vec{BE} = (x-1,y)$, so $\vec{EF} = (y,1-x)$, giving $F = E + \vec{EF} = (x+y, y+1-x)$.

Thus $\vec{DF} = F - D = (x+y, y+1-x) - (0, 1) = (x+y, y-x)$.

Hence $DF^2 = (x+y)^2 + (y-x)^2 = 2(x^2+y^2) = 2 AE^2$.

This is true wherever $E$ and $F$ are, and whatever the orientation of $BEFG$.

Answer 2

HINT: there's only one (two, if you count mirror reflections) orientation available for the inner square... though I wouldn't exactly call $E$ and $F$ "inside".

Answer 3

$(NB)^2+(EN)^2=(EB)^2=y^2$
$(NB)^2=y^2-(EN)^2$
$(x-AN)^2=y^2-(y\sin\theta)^2=(y\cos\theta)^2$
$x-AN=\pm y\cos\theta$
$AN=x-y\cos\theta(\because y,\cos\theta \text{ both positive})$
$(AE)^2=(AN)^2+(EN)^2$
$(AE)^2=(x-y\cos\theta)^2+(y\sin\theta)^2$
$(AE)^2=x^2-2xy\cos\theta+y^2\cos^2\theta+y^2\sin^2\theta=x^2-2xy\cos\theta+y^2$

$\sin(90-\theta)=\frac{EM}{y}, \therefore EM=y \cos\theta \therefore MN=y(\cos\theta+\sin\theta)$

$\therefore DP=x-y(\cos\theta+\sin\theta)$

$\cos(90-\theta)=\frac{MF}{y} \therefore MF=y\sin\theta \text{ and }PM=x-NB=x-y\cos\theta$

$PF=x-y(\cos\theta-\sin\theta)$

$(DF)^2=(DP)^2+(PF)^2$

Substitute and simplify to get $2(x^2+y^2-2xy\cos\theta)$

Hence $(DF)^2=2(AE)^2$

Answer 4

Let's solve the problem using complex numbers. WLOG, let $B$ be the origin and let $BADC$ and $BEFG$ be squares with points in that order counterclockwise. Let $z = BA$ and $w = BE$

Since $BC$ is just $BA$ rotated by $90^{\circ}$ CCW, we have $BC = i \cdot BA = iz$.

Then, since $BC = AD$, we have $BD = BA+AD = BA+BC = z+iz = (1+i)z$.

Similarly, $BG$ is just $BE$ rotated by $90^{\circ}$ CCW, so $BG = i \cdot BE = iw$.

Then, since $BG = EF$, we have $BF = BE+EF = BE+BG = w+iw = (1+i)w$.

Therefore, we have $|DF|^2 = |BF-BD|^2 = |(1+i)w-(1+i)z|^2 = |(1+i)(w-z)|^2$ $= |1+i|^2|w-z|^2 = 2|w-z|^2 = 2|BE-BA|^2 = 2|AE|^2$, as desired.

Note that this holds even if $E,F$ do not lie inside $ABCD$. Also, note that the above is a fancy way of showing that line segment $DF$ can be obtained by rotating line segment $AE$ about $B$ by $45^{\circ}$ and dilating by $\sqrt{2}$ about $B$.

Answer 5

∆ABE and ∆DBF are similar

(SAS similarity criterion).

BF/BE = √2.

The assertion follows.