Let K and L be circles with centres O and P, respectively, each of which is exterior to the other circle. Suppose K and L intersect at points A and D, and that the diameter AOB of K intersects L at a second point C lying between A and O. Extend line segment BD to meet L at a second point E.
(a) Prove that AE is a diameter of L. (b) Prove that OCPD is a cyclic quadrilateral.
I have tried to join A and D and assumed it is perpendicular to line BE. Also I have tried to use subtended angle, but I am stuck now since I am getting the same formula over and over. Also I know that I should be using Thales' theorem but I am not sure how to start.
First things first: by symmetry, $AD\perp OP$.
$B$ is the antipode of $A$ and $D$ is the symmetric of $A$ wrt the $OP$ line, so $BE\parallel OP\perp AD$.
This implies $\widehat{ADE}=90^\circ$, so $AE$ is a diameter of the second circle.
If $C$ is external to $AB$ (i.e. if $\widehat{BAE}$ is obtuse, like in the above diagram) $\widehat{OPD}=\frac{1}{2}\widehat{APD}=\widehat{ACD}=\widehat{OCD}$. This gives that $OCPD$ is cyclic.
In any case the circumcircle of $OPD$ is the nine point circle of $ABE$, since it goes through the midpoints $P,O$ and the projection of $A$ on $BE$. In particular it has to go through the projection of $E$ on $AB$, which is $C$ since $\widehat{ECA}=\widehat{EDA}=90^\circ$.