Prove that $M$ is the midpoint of $HG$

Question

In the image, $A,B$ and $E$ are tangency points;

$H,E,M$ and $G$ are collinear;

$O$ is the center of the circle whose diameter is $AB$

$OM \perp HG$

I did solve this problem with the following similarities $\triangle AHE \sim \triangle OME \sim \triangle BGE $ and some calculations, but I wonder if there is a simpler (more synthetic) way to do it.

It is fairly easy to see that $OM = \frac{BG-AH}2$

Answer 1

Let draw line CD through O perpendicular to MO. Triangles COB and DOA are right and equal by angle and hypotenuse. Then MG = OC = OD = MH.

Answer 2

Following Ivan Kaznacheyeu's comment:

Let $X = OG \cap AH$, then $\triangle XAO \cong GBO$ ($A-S-A$ on sides $OA=OB$).

So $OX=OG$

$\angle XHG = 90^{\circ} \implies HO=OG=OX \implies \triangle HMO \cong \triangle GMO \implies MH = MG$

Answer 3

The line $OM$ is an axis of symmetry of the circle. This line is parallel to $AH$ and $GB.$ Draw the circles symmetric to the given circles with respect to $OM.$

The rectangle with a diagonal $AB$ is inscribed in the large circle, with the aim to better visualize the situation:
$OM$ is the axis of symmetry of the rectangle, $HG$ is parallel to the adjacent side $AA',$ therefore $M$ is the midpoint of $HG.$

Answer 4

As can be seen in figure:

1-Regarding my comment, when $MO=0$ it means O coincides with M and B becomes B'; then triangles AHM and B'GM become congruent which in turn means GM=MH, that M is mid point of GH.

2- Another approach:

M is mid point of chord IJ, so we have:

$IM=JM\Rightarrow IG +GM=JH+HM$

The powers of points I and J related with circle center on O and passing points G and H' are equal, that is $IG=JH'$ which results in $H'M=MG$ and M is mid point of GH'. But H coincides on H' because of similarity of triangles EGB and MAH.