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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!

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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

$$O\in HG$$

Alternatively (simpler)

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 $O$.

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