<s>[INST] 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 $DF^2=2\cdot AE^2$.

 [/INST] 
$(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 $(DF)^2=2(AE)^2$
</s>