Parallel Chord problem

Since $AB$ is parallel to $CD$ and $ABCD$ is a cyclic quadrilateral, it must be a trapezoid. Draw $AF$ and $BE$ be parallel to $GH$ . Then $ABEF$ is a rectangle and $AB = EF$ . Since trapezoid $ABCD$ is symmetrical about its central axis, $CE=DF=4-3=\text{1 cm}$ . Therefore, $AB=EF=6-2=\text{4 cm}$ and the red line $AG=AB-GB=4-3=\boxed{\text{1 cm}}$ .