Prove that 3 circles are concurrent Given quadrilateral $ABCD$. $AB \cap CD = E, AD \cap BC = F$.  Prove the circles $(EF), (AC), (BD)$ are concurrent
 ($(XY)$ means circle with diameter $XY$)
 
 A: This result is famous enough to have a name.  It is called the Gauss-Bodenmiller Theorem.  It states that the circles you describe are coaxial.  That is, they share a common radical axis.  In this case, you have made the assumption that the three circles all intersect each other.  Then, the common radical axis is just the common chord, hence they are concurrent.
The key to proving this is to prove that the orthocenter of $\triangle BCE$ lies on each of the three radical axes between the three circles.  Let this orthocenter be $H$.
To prove that $H$ lies on the radical axis of $(FE)$ and $(AC)$, we actually bring in an auxillary circle $(EC)$.  We prove that $H$ lies on the radical axis between $(EF)$ and $(EC)$, and between $(EC)$ and $(AC)$.  Both of these claims are proven by observing that since these are circles whose diameters are a side of a triangle, they will share some common altitude, that $H$ lies on.  
After this, we have by the radical axis concurrency theorem that $H$ lies on the radical axis between $(FE)$ and $(AC)$.  We repeat a similar procedure for each of the other two pairs.
The Theorem has a second statement that the orthocenters of $\triangle FAB$, $\triangle FDC$, $\triangle EAD$, and $\triangle BCE$ all lie on this common radical axis of the three circles.  This makes this line cool, so it is called the Steiner line.
(All information in this answer was stolen from Evan Chen's Euclidean Geometry in Mathematical Olympiads.  You should buy it!)
