As has been shown in the figure below: $AB$ is the diameter of $\odot O$, $CE$ and $CF$ are tangent lines to $\odot O$, and $D$ is the intersection point of $AE$ and $BF$.
Show that the $CD$ is perpendicular to $AB$.
This is easy to prove by analytical method. I am wondering whether there is any pure geometric method.
updated figure
I'm a bit tired, so I am going to cheat the system and take out the big guns...
Let $t_E$ and $t_F$ be the tangent lines to the circle $(O)$ at the points $E$ and $F$ respectively and let $K$ be the intersection point of the lines $AF$ and $BE$.
Now, consider the (degenerate) hexagon $AEEBFF$, inscribed in the circle $(O)$. By Pascal's theorem, the intersection point $C = t_E \cap t_F$, the intersection point $D = AE \cap BF$ and the intersection point $K = AF \cap BE$ are collinear.
Since $AB$ is a diameter, $\angle \, AEB = \angle \, AFB = 90^{\circ}$. Therefore $BF$ and $AE$ are two out of the three altitudes of triangle $ABK$ and their intersection point $D$ is the orthocenter of $ABK$. Consequently, the line $KD$ is the third altitude of the triangle and is therefore orthogonal to $AB$. However, it was proved earlier that $C$ is collinear with $K$ and $D$, i.e. $C$ lies on $KD$. Therefore $CD$ is orthogonal to $AB$.