let $KS$ be $\parallel$ $MA$ with $S$ on the circle. easily,$MK=AS$ so it suffices to show that the perpendicular from $O$ to $AS$ is half the hypotenuse$(AC)$. since, $KS$ $\parallel$ $MA$ $\angle SKC=90^\circ$ and so $SC$ is a diameter and $O$ lies on $SC$. therefore $\triangle CAS$ is also right angled and therefore the perpendicular from $O$ to $AS$ is $\parallel AC$ and passing through the midpoint of $SC(O)$ and by the midpoint theorem is half of $AC$.

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Let the right triangle be $ABC$ with right angle at $C$ and $M$ on $AC$, and let $P,N$ be the midpoints of $AB$ and $MK$, respectively. By PoP, $\frac{CM}{CK}=\frac{CB}{CA}$, so $CN$ is isogonal to $CP$. Since $\angle PCB=\angle B$, this means that $CN\perp AB$, and similarly $CP\perp MK$. But since $OP\perp AB$ and $ON\perp MK$, this means $CNOP$ is a parallelogram, so $ON=CP=\frac{1}{2}AB$.