Geometry ... (1)

$\begin{aligned}& a^2+b^2+c^2+d^2=ab+bc+cd+da \\& \implies a^2+b^2+c^2+d^2-ab-bc-cd-da=0\\& \\&\color{#3D99F6}{\text{multiply both sides with '2'}}\\& \\&\implies 2a^2+2b^2+2c^2+2d^2-2ab-2bc-2cd-2da=0\\& \implies (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0 \end{aligned}$

$\therefore a=b=c=d$ . Hence, $\text{quadrilateral} ~ABCD ~\text{is a}~ \color{#D61F06}{\text{RHOMBUS}}$

$[ABCD]=\dfrac{pq}{2}$ where $p$ and $q$ are the diagonals of rhombus.

Since $[ABCD]=60, ~p=30$

$\therefore \begin{aligned}&60=\dfrac{30}{2}q \\& \implies q=4\end{aligned}$

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NOTE: $[\cdot]$ represent the area of the polygon