5. In the figure below, $A B C$ is an isosceles triangle inscribed in a circle with centre $O$ and diameter $A D$, with $A B=A C$. $A D$ intersects $B C$ at $E$, and $F$ is the midpoint in $\mathrm{cm}$. of $O E$. Given that $B D$ is parallel to $F C$ and $B C=2 \sqrt{5} \mathrm{~cm}$, find the length of $C D$
(A) $\frac{3 \sqrt{5}}{2}$
(B) $\sqrt{6}$
(C) $2 \sqrt{3}$
(D) $\sqrt{7}$
(E) $2 \sqrt{6}$
5. Answer: (B)
Since the diameter $A D$ perpendicularly bisects the chord $B C$,
$$
B E=E C=\sqrt{5} .
$$

Also, given that $B D / / F C$, we have $\angle D B E=\angle F C E$. Thus $\triangle B D E$ is congruent to $\triangle C F E$, so $D E=F E$. As $F$ is the midpoint of $O E$, we have $O F=F E=E D$.
Let $O F=x$. Then $A E=5 x$.
Using Intersection Chord Theorem, we have
$$
A E \times E D=B E \times E C,
$$
which leads to $5 x^{2}=5$. Consequently we obtain $x=1$. Now $C D^{2}=C E^{2}+E D^{2}$ gives $C D=\sqrt{5+1}=\sqrt{6}$.