Remember it's cyclic

Since $ANBM$ is a cyclic quadrilateral , the opposite angles $\angle ANB + \angle AMB = 180^\circ$ $\implies \angle AMB = 60^\circ$ . Also $\angle MBA = MNA = 60^\circ$ and $\angle MAB = \angle MNB = 60^\circ$ . Therefore $\triangle ABM$ is equilateral and $AB=BM=MA = c$ .

Let the length of diagonal $MN$ be $x$ . Then applying Ptolemy's theorem : the product of the lengths of the two diagonals of a cyclic quadrilateral is equal to the sum of the products of opposite sides, we have $xc = ac + bc$ $\implies MN = x = \boxed{a+b}$ .

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