Competition style problem circa 1992 We're given a triangle $ABC$.
Going clockwise, let $B_1$ and $B_2$ be distinct points on the segment $AC$ ($B_1$ is between $A$ and $B_2$), let $A_1$ and $A_2$ be distinct points on the segment $CB$ ($A_1$ is between $C$ and $A_2$), and finally let $C_1$ and $C_2$ be distinct points on the segment $AB$ ($C_1$ is between $B$ and $C_2$).
The circumcircles of triangles $CB_1A_1$ and $CB_2A_2$ intersect at $C_3 \neq C$.
The circumcircles of triangles $BA_1C_1$ and $BA_2C_2$ intersect at $B_3 \neq B$.
The circumcircles of triangles $AB_1C_1$ and $AB_2C_2$ intersect at $A_3 \neq A$.
Prove that the lines $AA_3$, $BB_3$ and $CC_3$ have a common point.

Consider the region near vertex $A$. (Point $B$ on ray $AC_2$ and point $C$ on ray $AB_1$ not pictured.) 


$\square AB_1A_3C_1$ is cyclic, so opposite angles $\angle B_1$ and $\angle C_1$ are supplementary; consequently, $\angle B_1 \cong \angle A_3 C_1 C_2$ (the latter itself being a supplement of $\angle C_1$). Likewise, $\angle C_2 \cong \angle A_3 B_2 B_1$.

By the Angle-Angle Similarity Theorem, we have that $\triangle A_3 B_1 B_2 \sim \triangle A_3 C_1 C_2$, and we deduce that the "bases" ($B_1B_2$ and $C_1C_2$) and corresponding "altitudes" ($A_3P$ and $A_3Q$) of these triangles have proportional lengths; but note that those altitudes are directly related to the angles into which $\angle A$ has been divided. Therefore,
$$\frac{|B_1B_2|}{|C_1C_2|} = \frac{|A_3P|}{|A_3Q|} = \frac{|AA_3|\sin\angle A_3AB_1}{|AA_3|\sin\angle A_3AC_2} = \frac{\sin\angle A_3AC}{\sin\angle A_3AB}$$
Likewise for vertex $B$ and vertex $C$, so that the trigonometric form of Ceva's Theorem is satisfied:
$$
\frac{\sin\angle A_3AC}{\sin\angle A_3AB} \;
\frac{\sin\angle B_3BA}{\sin\angle B_3BC} \;
\frac{\sin\angle C_3CB}{\sin\angle C_3CA} \;
= \;
\frac{|B_1B_2|}{|C_1C_2|} \;
\frac{|C_1C_2|}{|A_1A_2|} \;
\frac{|A_1A_2|}{|B_1B_2|} \;
= \; 1$$
proving that the lines $AA_3$, $BB_3$, $CC_3$ share a common point.
