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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.

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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?

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  • $\begingroup$ 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? $\endgroup$
    – James Ko
    Commented Sep 29, 2017 at 3:55
  • $\begingroup$ Oh! I see, that's why $CD = AB$ is useful... $\endgroup$
    – James Ko
    Commented Sep 29, 2017 at 3:55
  • $\begingroup$ 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? $\endgroup$
    – James Ko
    Commented Sep 29, 2017 at 4:00
  • $\begingroup$ You have $1/x = (1+x)/1.$ Have you done the Golden Ratio? $\endgroup$
    – David K
    Commented Sep 29, 2017 at 11:47

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