Prove that the Circumcentre, Centroid, and Orthocentre are collinear in triangle $\triangle ABC$ if $\angle BAC >90^{\circ}$

Question

Let the altitudes of the obtuse triangle $\triangle ABC$, where $\angle BAC >90^{\circ}$, meet at orthocentre $H$. The circumcentre of $\triangle ABC$ is $O$. Let $M$ be the midpoint of $BC$, and let $G$ be the centroid. Additionally, $Y$ is the reflection of $H$ over $M$.

1. Show that $AY$ is the diameter
2. Show that $AH=2OM$
3. Hence show taht $O,H,G$ are collinear

The similarity of trianges $\triangle AGH$ and $\triangle MGO$ (by SAS) implies that $HA\parallel MO$ so line $HA$ is the altitude of $\triangle ABC$. A similar proof can be made for $B$ and $C$. Another way to prove this is to use the medial triangle where the circumcenter of $ABC$ is the orthocentre of the medial triangle and the centroid of $ABC$ is the centroid of the medial triangle, when scaled.

Could someone please provide some help regarding how we should prove this obtuse case acknowledging the differences to the acute case for $\triangle ABC$?

All help is appreciated, thank you.

Answer 1

1- We use this property that the circle d passing through vertexes B and C and orthocenter H is congruent with the circumcircle c of triangle ABC.So k is the reflection of H over BC and Z is the reflection of Y over BC. Point M is also the midpoint of JR, where J and R are the intersections of HK and YZ with BC respectively. This means common chords HK and YZ are in the same distance from MO which is the perpendicular bisector BC. , that I they are equal tnd the qudrilateral HKYZ is a rectangle,so we have:

$\angle HKY=\angle AKY=90^o$

This means AY is the diameter of the circumcircle c.

2- We use this fact that the nine point circle e passes through the midpoint N of AH.In triangle AHY, N is the midpoint of AH and O is the midpoint of AY, so we have:

$NO||HM$

Also :

$MO||AH$

because they are both perpendicular to BC, hence quadrilateral HNOM is a parallelogram and we have:

$MO=HN=\frac 12 AH$

3- As can be seen in the picture OH in indeed the diagonal of the parallelogram HNOM, Also AM is the medians of triangle AHY because the construction is symmetric over point M which means $HM=YM$ and HO is another median of triangle AHY which must intersect at G. This must be explained more. In this way H, O and G are colinear.