Define $E$ as the intersection of $(O_1)$ and $(O_2)$. By angle chasing we can prove that $AEOB$ and $DEOC$ are cyclic quadrilaterals. Therefore, using power of points, then $AB, CD, OE$ concurrent at $F$. Similarly, $AD, HE, BC$ concurrent at $G$. Using Brocard theorem, we get $OE \perp HE$. But $O_1O_2 \perp HE$ so $O_1O_2 \parallel OE$. Therefore $O_1O_2$ passes through the midpoint of $OH$ labeled $I$. Now we need to prove that $O(EHO_1O_2)= -1$, which is equivalent of proving $G(HFDC)= -1$, which is obviously true. Therefore $I$ is the midpoint of $O_1O_2$. From there we conclude that $HOO_1O_2$ is a parallelogram, which leads to $\angle HO_1O= \angle HO_2O$ or $\angle ON_1N_2= \angle ON_2N_1$, leads to $ON_1= ON_2$. After that it's easy to conclude that $M_1N_1= M_2N_2$ and we're done.

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Without loss of generality treat $M_1$, $N_1$, $N_2$, $M_2$ are in this order. Since $OM_1 = OM_2$ and $\angle OM_1N_1 = \angle OM_1M_2 = \angle OM_2M_1 = \angle OM_2N_2$, it suffices to prove $ON_1 = ON_2$ - it would then follow that $\angle M_1N_1O = 180^{\circ} - \angle ON_1N_2 = 180^{\circ} - \angle ON_2N_1 = \angle ON_2M_2$ and we would have $\triangle OM_1N_1 \cong \triangle OM_2N_2$. The Sine Law gives $\frac{ON_1}{\sin \angle N_1HO} = 2R_{OHO_1}$ and $\frac{ON_2}{\sin \angle N_2HO} = 2R_{OHO_2}$ and since the two sines are equal (the angles in them are adjacent), we reduce to proving $R_{OHO_1} = R_{OHO_2}$. Since $OH = 2R_{OHO_1}\sin \angle OO_1H$ and $OH = 2R_{OHO_2}\sin \angle OO_2H$, it now remains to show $\angle OO_1H = \angle OO_2H$. If we denote $\angle HAD = \angle HBC = \alpha$ and $\angle HDA = \angle HCB = \gamma$, then $\angle O_1HD = 90^{\circ} - \frac{\angle HO_1D}{2} = 90^{\circ}-\alpha$ and so for $K = O_1H \cap AD$ we get $\angle DKH = 90^{\circ} + \alpha - \gamma$, so $\angle OO1H = \gamma - \alpha$. Analogously $\angle OO_2H = \gamma - \alpha$ and so we are done.

Today I realized that this problem has a very straightforward angle chase solution! Without loss of generality treat $M_1$, $N_1$, $N_2$, $M_2$ are in this order. We firstly eliminate $M_1$ and $M_2$ from the diagram. Since $OM_1 = OM_2$ and $\angle OM_1N_1 = \angle OM_1M_2 = \angle OM_2M_1 = \angle OM_2N_2$, it suffices to prove $ON_1 = ON_2$ -- it would then follow that $\angle M_1N_1O = 180^{\circ} - \angle ON_1N_2 = 180^{\circ} - \angle ON_2N_1 = \angle ON_2M_2$ and we would have $\triangle OM_1N_1 \cong \triangle OM_2N_2$. Next, we eliminate $N_1$ and $N_2$ from the diagram. Note that $ON_1 = ON_2$ is equivalent to $\angle ON_1H = \angle ON_2H$ and from the circles the latter is as showing $\angle OO_1H = \angle OO_2H$. If we denote $\angle HAD = \angle HBC = \alpha$ and $\angle HDA = \angle HCB = \gamma$, then $\angle O_1HD = 90^{\circ} - \frac{\angle HO_1D}{2} = 90^{\circ}-\alpha$ and so for $K = O_1H \cap AD$ we get $\angle DKH = 90^{\circ} + \alpha - \gamma$, so $\angle OO1H = \gamma - \alpha$. Analogously $\angle OO_2H = \gamma - \alpha$ and so we are done.