Let $A,B,C,D$ lie on a circle, $AD$ and $BC$ meet at $E$. The circle through $A,C,E$ have centre $O$. Prove that $BD$ is perpendicular to $EO$.

Question

Let A,B,C,D lie on a circle, where AD and BC meet at E. Let the circle through A,C,E have centre O. Prove that BD is perpendicular to EO.

I saw this question posted earlier but it was put on hold and now I can't find it anymore... I thought I'd repost with my solution. Not sure about the rules on that; Let me know if I have to take it down.

Are there any other ways to prove this? There was an answer to the old post regarding defining a circle through B,E,G, with BE as the diameter, and by Thales theorem this would prove that the angle at G is 90 degrees. But how can you prove that BE would be the diameter of the circle through B,E,G without assuming angleBGE to be 90 degrees?

Picture by someone who answered the original post

Answer 1

I used circle theorems (which I just realised I forgot to name exactly).

Answer 2

You can also use Reim's theorem:

https://mathsolympian.wordpress.com/2014/08/02/reims-theorem/

Then by Reim we have that tangent $EE$ on circle $ACE$ is parallel to $DB$. But since tangent $EE$ is perpendicular to radius $EO$ we are done.