Chord $AB$ of a circle is extended through $B$ to an exterior point $D$, and a line is drawn from $D$ tangent to the circle at $C$. If $AB$ = $CD$ and triangle $BCD$ has area $2 \sqrt 5$ square centimeters, then the area of triangle ABC can be expressed as $a + b \sqrt 5$ square centimeters. Find the ordered pair $(a, b)$.
What I tried: Using the Angle of the Alternate Segment Thm. to show $\angle {BCD} \cong \angle {CAB}$, concluding $\triangle {BCD} \simeq \triangle {ACD}$ (they also share an angle at $D$). I then tried to subtract the area of $\triangle {BCD}$ from $\triangle {ACD}$ hoping I would find something that made everything cancel out, but to no avail.
I have a strong feeling I'm supposed to use power of a point, but I'm not sure how to use it effectively.
Use the power of a point to find the ratio $AB:AD$ and hence the ratio $AB:BD.$
You now have two triangles, each of which has $C$ as one vertex while the opposite side is on line $AD.$
Can you finish it now?