Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let the diagonals $AC,BD$ intersect at $E$. 

Let $\theta = \angle AEB$ and $t=\cos\theta$. So, $\cos(180^\circ-\theta)=-\cos\theta=-t$. 

Using the cosine rule:

$AB^2 = AE^2+BE^2-2AE\cdot BE\cdot \cos\theta=AE^2+BE^2-2AE\cdot BE\cdot t$

$CD^2 = CE^2+DE^2-2CE\cdot DE\cdot \cos\theta= CE^2+DE^2-2CE\cdot DE\cdot t$

$BC^2 = CE^2+BE^2-2BE\cdot CE\cdot \cos(180^\circ-\theta)= CE^2+BE^2+2BE\cdot CE\cdot t$

$AD^2 = AE^2+DE^2-2AE\cdot DE\cdot \cos(180^\circ-\theta)= AE^2+DE^2+2AE\cdot DE\cdot t$

$(1)+(2)-(3)-(4)\implies$

$AB^2+CD^2-AD^2-BC^2=-2AE\cdot BE\cdot t-2CE\cdot DE\cdot t-2BE\cdot CE\cdot t-2AE\cdot DE\cdot t$

$\implies 0 = -2t(AE\cdot BE+AE\cdot DE+CE\cdot BE+CE\cdot DE)$

$\implies t(AE+CE)(BE+DE)=0 $

$\implies t\cdot AC\cdot=0BD$

But $AC,BD$ cannot be $0$. So, $t=0\implies \cos\theta = 0\implies \theta = 90^\circ$

So, $AC$ and $BD$ are perpendicular to each other.

Cosine Rule

The cosine rule states that in a triangle $ABC$, $AB^2=AC^2+BC^2-2BC\cdot AC\cdot \cos \angle ACB$