For reference: Inside a regular hexagon $ABCDEF$ mark a point $P$ such that $AP=a$, $PC=b$ and the measure of the $\angle APC = \angle AFD$. Calculate the smallest value of $EF$.(Answer: $\sqrt{a^2+b^2}$)
My progress:
Draw the figure we have
$\triangle APC: a^2+b^2 =AC^2 \implies AC = \sqrt{a^2+b^2} = DF$
$\triangle FEG:cos30^o = \frac{FG}{L} \implies \frac{\sqrt3}{2}=\frac{\sqrt{a^2+b^2}}{2L}\\ \therefore L =\frac{\sqrt3\sqrt{a^2+b^2}}{3} $
but what would be the condition to have the value of the minimum side?