### 2003 CentroAmerican #2

$S$ is a circle with $AB$ a diameter and $t$ is the tangent line to $S$ at $B$. Consider the two points $C$ and $D$ on $t$ such that $B$ is between $C$ and $D$. Suppose $E$ and $F$ are the intersections of $S$ with $AC$ and $AD$ and $G$ and $H$ are the intersections of $S$ with $CF$ and $DE$. Show that $AH=AG$.

Note $\angle CDF=90^{\circ}-\angle FBD=\angle ABF=\angle AEF=180^{\circ}-\angle FEC$, hence $CEFD$ is cyclic. Thus $\angle AEH=180^{\circ}-\angle DEC=180^{\circ}-\angle DFC=\angle GFA$, so arc $AG$ and arc $AH$ have the same length, hence $AG=AH$, as desired. $\square$