Problem involving three cyclic quadrilaterals

Question

I found out the following configuration in elementary geometry. I know it is true (by drawing in Geogebra) but I haven't proved it yet.

Let $ABC$ be a triangle with the circumcircle $(O)$. $M$ is an arbitrary point on the edge $BC$. $AM$ intersects $(O)$ at $D$. The circumcircle of triangles $CMD$ and $BMD$ intersects the line $AC$ and $AB$ at $E$ and $F$. Then

• $AEDF$ is a cyclic quadrilateral.
• Let $H, I, G$ be the circumcenter of the circumcircle of three triangles $BMD$, $CMD$, $AEF$ correspondingly. Then five points $G, O, H, D, I$ are concyclic.

My question is: Is it a familiar configuration or result in elementary geometry? Are there any source to check new result in elementary geometry?

Please share with me more information about it.

Thank you.

Answer 1

$$\measuredangle DEC=\measuredangle DMC=\measuredangle BFD,$$ which says that $AEDF$ is cyclic.