Parallelogram and circumscribed circle

We have a parallelogram $$ABCD$$, $$\angle{A}<90^\circ$$ and its diagonals intersect at $$O$$. $$DH$$ - height, $$H\in$$ AB and $$k(Q; r): B \in k, D \in k \text{ and } H \in k$$. (a circle around the triangle BDH). $$k$$ intersects $$CD$$ at $$G$$. We have to show that the line $$HG$$ passes through $$O$$.

I have noticed that OH = OD = OB = OG but I use that O matches with Q. How can I proof that? I'm not sure that you understand me but that's because of my English. I'm really sorry and if there is anything that you don't understand I will be very pleased to clear it. Thank you in advance!

By the converse of Thales' Theorem, $$DB$$ is the diameter of $$k$$ and $$O$$ its center.
Let $$t$$ be the line parallel to $$DH$$ through $$O$$. Since $$O\in t$$ and $$HB, DG\perp t$$, we notice that $$t$$ is a symmetry axis. Thus $$G=D'$$ and $$B=H'$$.
However, this also means that the line $$HG$$ is the reflection of $$DB$$. They obviously meet at the symmetry axis, specifically at $$O$$ since $$O\in DB$$. Thus
$$O\in HG$$
Observe that $$\angle DGH=180°-\angle BHD=90°$$ since $$DHBG$$ is cyclic. Since $$DG\parallel HB$$, you have that $$\angle DGB=\angle GBH=90°$$ Thus - by the converse of Thale's Theorem - $$HG$$ is a diameter and therefore contains the center $$O$$.