Two overlapping squares $ABCD$ 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 $DF^2=2\cdot AE^2$.

 A: 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 $BEFG$.
A: 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".
A: 
$(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 $(DF)^2=2(AE)^2$
A: 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 $B$.
A: ∆ABE and ∆DBF are similar
(SAS similarity criterion).
BF/BE = √2.
The assertion follows.
