Raja Oktovin wrote: Let $ C_1$ be a circle and $ P$ be a fixed point outside the circle $ C_1$. Quadrilateral $ ABCD$ lies on the circle $ C_1$ such that rays $ AB$ and $ CD$ intersect at $ P$. Let $ E$ be the intersection of $ AC$ and $ BD$. (a) Prove that the circumcircle of triangle $ ADE$ and the circumcircle of triangle $ BEC$ pass through a fixed point. (b) Find the the locus of point $ E$. a, Let $ PK,PQ$ are two tangents from P to $ (C_1), PO\cap KQ = \{H\}, CH\cap (C_1) = \{J\}, BH\cap (C_1) = \{F\}$ We have $ HA.HO = HK.HQ = HC.HJ$ so PJOC is cyclic we get $ \angle OJH = \angle OPC (1)$ On the other hand, $ OH.OP = R^2 = OD^2$ then $ \frac {OH}{OD} = \frac {OD}{OA}$ therefore $ \Delta HOD$~$ \Delta DOP$ then $ \angle OPC = \angle ODH (2)$ From (1) and (2) we have $ \angle HJO = \angle HDO$ so $ HO\perp JD$ therefore $ JD//KQ$ Similarly, $ AF//KQ//JD$ then $ arcAD = arcJF$ We get $ \angle BHC = \angle BEC$ then BHEC is cyclic, similarly AHED is cyclic So the circumcircle of triangle $ ADE$ and the circumcircle of triangle $ BEC$ pass through a fixed point H b, We have $ 180^o - \angle BHE = \angle BCE = 1/2(arc BK + arc KA) = 1/2(arc BK + arc QF)$ $ = \angle BHK$ Then K,H,E is collinear So $ E\in [KQ]$

Let $O$ be the center of $C_1$ and $\tau$ be the polar of $P$ WRT $C_1$ (Locus of E), $F \equiv OP \cap \tau.$ Denote $\omega_1,\omega_2$ the circumcircles of $\triangle EBC ,\triangle PBC,$ respectively. Line $AD \equiv a$ is both image of $\omega_1$ and $\omega_2$ through the inversions with poles $E,P$ and powers $p(E,C_1),$ $p(P,C_1),$ respectively. Hence, by conformity $\angle (a,C_1)=\angle (\omega_1,C_1)=\angle (\omega_2,C_1).$ Since $B,C$ are double points in the inversion $\mathcal{I}$ about $C_1(O),$ then $\omega_1$ and $\omega_2$ are homologous under $\mathcal{I}.$ Keeping in mind that $\mathcal{I}: F \mapsto P,$ it follows that $F \in \odot(BEC).$ By analogous reasoning we get that $F \in \odot(AED).$