# Deductive Geometry - Cyclic Quadrilaterals Help

• September 24th 2009, 03:54 AM
BG5965
[Solved] Deductive Geometry - Cyclic Quadrilaterals Help
Hi, here is the question and diagram.

http://i301.photobucket.com/albums/n...oether14-1.png

I know that OECD is a cyclic quadrilateral, but I'm not sure how to work from there.

Thanks for helping, BG
• September 24th 2009, 04:52 AM
CaptainBlack
Quote:

Originally Posted by BG5965
Hi, here is the question and diagram.

http://i301.photobucket.com/albums/n...oether14-1.png

I know that OECD is a cyclic quadrilateral, but I'm not sure how to work from there.

Thanks for helping, BG

$\angle AEO= \angle OAE$ as $\triangle AOE$ is equilateral

This implies that $\angle DEB = \angle EBD$ since $\angle CAB+\angle ABC= 90^{\circ}$ and $\angle AEO+ \angle DEB=90^{\circ}$.

CB