We have $\angle{AO_2C}=360-2\angle{CFA}=360-2(180-\angle{A})=2\angle{A}$. Combining this with $AO_2=O_2C$, we have $\triangle{AO_2C}\sim \triangle{BOC}$. From this we obtain $AO_2=\frac{b}{a}R$. Similarly from $\triangle{AO_1B}\sim \triangle{BOC}$, we obtain $AO_1=\frac{c}{a}R$. Since $\triangle{OO_2A}\cong \triangle{OO_2C}$, we have $\angle{OAO_2}=\angle{O_2CO}=\angle{C}$. Thus by the law of sines in $\triangle{AOO_2}$, we have $OO_2=\dfrac{\sin{\angle{C}}}{\sin{\angle{B}}}AO_2=\frac{c}{a}R=AO_1$. Similarly we have $OO_1=AO_2$. Therefore $AO_1OO_2$ is a parallelogram which gives us our desired result.

$AO_1OO_2$ is a $|| gm$ and thus midpoint $M$ of $AO$ must lie on $O_1O_2$.

Hint

Pretty nice problem. From angle chasing we have ∠AEB=∠AFC=180-∠BAC. Then ∠AO1B=∠AO2C=2×∠BAC. Then ∠O1AB=∠O2AB=90-∠BAC so we have that O1A is perpendicular to AC and O1A||OO2 (because points O, X, O2, where X is midpoint of AC are collinear by definition of points O and O2, so OO2 is perpendicular bisector of AC). Similary OO2||AO1. AOO1O2 is a parallelogram so we are done.

Consider the configuration in the picture in the exam paper. One can use directed angles to fix it if it is another configuration, but we leave it as an exercise to the reader. $\angle EFO=90^{\circ}-(90^{\circ}-A)=A=\angle FEO.$ We now prove $\angle AO_2O$ is $A.$ Indeed, $2\angle AO_2O=\angle AOC=2A$ by congruent triangles. Thus, we now now know that $AFOO_2$ is cyclic. $\angle O_1AM=\angle O_1AO=\angle O_1BO=90^{\circ}-C+\angle ABO_1=90^{\circ}-C+\angle AEB-90^{\circ}=\angle AEB-C=B.$ This is very helpful, since similarly $\angle O_2AM=C.$ We now know $\angle O_1AO_2=180^{\circ}=A.$ Now, we shall finish by noting $\angle AO_2O=\angle EFO=A,$ thus $\angle O_1AO_2+\angle AO_2O=180^{\circ},$ so $O_1A\parallel OO_2,$ and similarly $O_2A\parallel OO_1,$ thus $M$ is the center of parallelogram $AO_1OO_2,$ thus it lies on $O_1O_2.$