It follows from the well-known lemma that "For any $\triangle{ABC}$ with orthocenter $H$ and $M$ be the midpoint of $BC$, if $H’$ is the reflection of $H$ over $M$, then $AH'$ is diameter of $(ABC)$.” Applying the lemma for $\triangle{HBC}$, we get that $AM\cap HO_2=T\in (HBC)$. So, $O_2$ is the midpoint of $HT$ and $O_1$ is the midpoint of $HA$. We have $O_1O_2\parallel AM$.

Let the circumradius of $\triangle ABC$ be $R$. Lemma 1: The circumradius of $\triangle BHC$ is $R$. Proof: Note that $\angle BHC=180^{\circ}-\angle A$ so $\sin\angle BHC=\sin\angle A$. Hence, $\frac{BC}{\sin\angle BHC}=\frac{BC}{\sin\angle A}=2R$, which proves the Lemma. $\Box$ Note that $\angle BO_2C=2\angle A$. By Lemma 1, $O_2B=O_2C=R$ and $\angle O_2BC=90^{\circ}-\angle A$. Hence, $MO_2=R\cos A$. Lemma 2: $AH=2R\cos A$. Proof: Let $B'$ and $C'$ be on $AC, AB$ respectively such that $BB'\perp AC$ and $CC'\perp AB$. Since $BC'B'C$ is cyclic, $\angle AC'B'=\angle C$ so $\triangle AC'B'\sim \triangle ACB \implies \frac{AH}{2R}=\frac{AC'}{AC}=\cos A$ (ratios of diameters of circumcircles=ratios of sides). $\Box$ By Lemma 2, $AO_1=R\cos A$. Hence, $AO_1=MO_2$ and $MO_2\parallel AO_1$ so $O_1AMO_2$ is a parallelogram. $\blacksquare$

Dear Mathlinkers, according to Carnot, AH = 2.MO2 and we are done... Sincerely Jean-Louis

HM-point,it could be a smart and easy solution.

Mx.22 wrote: HM-point,it could be a smart and easy solution. Nice observation!

$O_2$ is the reflection of $O$ to $BC$ so $O_2M=OM=\frac{AH}{2}=AO_1$ and $AO_1 \parallel O^2M$ because $AH _\perp BC\perp O_2M$

Really easy with nine-point circle properties. It suffices to show that $AO_2$ and $O_1M$ bisect each other but just note that $A$ is the orthocenter of $\triangle BCH$ and that $O_1M$ is the diameter of the nine-point circle of $\triangle BCH$. It's well known that the circumcenter of the nine-point circle is the midpoint of the line that joins the circumcenter and the orthocenter, in this case $AO_2$, and we're done.

Observe that $A$ is the orthocentre of $BHC$, so$$AO_1=\frac{AH}{2}=\frac{2O_2M}{2}=O_2M$$and $AO_1\parallel AH\perp BC\perp O_2M$. Hence $AO_1=O_2M$ and $AO_1\parallel O_2M$. Done.

Use complex numbers with $(ABC)$ as the unit circle. We have $o_2=b+c$ because $(b+c)-b=c$, $(b+c)-c=b$, $(b+c)-h=-a$ and all of them have modulus equal to $1$. Hence$$a+o_2=a+b+c=\frac{a+(a+b+c)}{2}+\frac{b+c}{2}=h_1+m,$$so $O_1AMO_2$ is a parallelogram. Done.

Note that $AO_1$ and $MO_2$ are both perpendicular to $BC$ and thus parallel. Further we have $AO_1 = MO_2 = R\cos A$, so we're done. jayme wrote: Dear Mathlinkers, according to Carnot, AH = 2.MO2 and we are done... Sincerely Jean-Louis What is Carnot?