# How to solve this circle geometry problem using power of a point?

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.

## 1 Answer

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?

• I'm confused. Wouldn't using power of a point wrt $D$ get me $CD^2 = BD \cdot AD$? How would I get the ratio $AB : AD$ from that? – James Ko Sep 29 '17 at 3:55
• Oh! I see, that's why $CD = AB$ is useful... – James Ko Sep 29 '17 at 3:55
• David, what I have now is $[ABC] : [BCD] = AB : BD = AD : AB$. I feel really close because I know $[BCD] = 2 \sqrt 5$, but what do I do now? – James Ko Sep 29 '17 at 4:00
• You have $1/x = (1+x)/1.$ Have you done the Golden Ratio? – David K Sep 29 '17 at 11:47