Show that tangents to circle are concurrent using cross-ratios. Can anyone give me some direction? I do not know why the whole cross-ratio thing throws me off so much.
Let $ABCD$ be a quadrilateral inscribed in a circle.  If $U$ is the intersection of its diagonals and $V$ is the intersection of sides $AB$ and $CD$, show that the tangents to the circle at $A$ and at $D$ meet line $UV$ as the same point.  Assume that neither of these tangents is parallel to $UV$.
Hint: Let $X$ and $Y$ be the points where the two tangents meet $UV$ and show that $\textbf{cr}\left(P, U, Q, X \right) = \textbf{cr}\left(P, V, Q, R \right) = \textbf{cr}\left(P, U, Q, Y \right)$, where $P$ and $Q$ are the points where line $UV$ crosses the circle and $R$ is the intersection of side $AD$ with $UV$, as in the diagram.
Note: Similar reasoning shows that the tangents at $B$ and $D$ also intersect at a point of line $UV$.
*I would really appreciate any help I can get. I get mixed up about how $\textbf{cr}\left(P, V, Q, R \right) $ is supposed to fit in all this. I do not know how to show it to be equivalent to the other two which are more obviously related. Also, how does this lead to $X = Y$? 
 A: The key here is that the cross ratio between four collinear points is the same as the cross ratio between four lines connecting these points to a fifth point. I'll call the latter the cross ratio of four points seen from a fifth point.
Take the cross ratio $\operatorname{cr}(P,U,Q,X)$ seen from $A$. That's the cross ratio between the lines $(PA,UA,QA,XA)$ or equivalently $(PA,CA,QA,AA)$ where $AA$ denotes the limit case where two points on the circle are brought to become the same, in which case the secant joining them becomes the tangent. Now the circle is a special kind of conic, and the cross ratio of four points on a conic is the same no matter from which other point on the conic it is seen. So instead of the cross ratio $(P,C,Q,A)$ seen from $A$ you could consider $(P,C,Q,A)$ seen from $D$, i.e. $(DP,DC,DQ,DA)$ or equivalently $(DP,DV,DQ,DR)$. Now since $(P,V,Q,R)$ are collinear their cross ratio is the same no matter from which point you see it. This gives your first equality, the second is obtained in an analogous way.
If two cross ratios are equal, and share three points, then the fourth point has to be the same for both as well, since the cross ratio with respect to three distinct points uniquely identifies any point on a projective line. Therefore $X$ and $Y$ must be the same if they have the same cross ratio with respect to $P,U,Q$.
