# Geometry Problem — Find the area of the circle

Points $A, B, C, D$ are on a circle such that $AB = 10$ and $CD = 7$. If $AB$ and $CD$ are extended past $B$ and $C$, respectively, they meet at $P$ outside the circle. Given that $BP = 8$ and $∠AP D = 60º$, ﬁnd the area of the circle.

Based on the information, I came up with the following sketch:

Based, on the given info, and the theorem of geometry that states that the product of two secants and their external parts are equal to each other ($AP\cdot BP\; =\; \mbox{C}P\cdot DP$) I was able to find that $DP = 9$.

However, after this point I am stuck. I know I need to somehow find the radius, but I don't know how to proceed.

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I got that you can find $AC$ by law of cosines. However, how did you get that $∠AOC=60º$? If you are going by the inscribed angle theorem, $∠APD$ isn't an inscribed angle though. EDIT: The comment I responded to was removed... – 1110101001 Dec 21 '13 at 20:36

Hint:

$PD=AP/2 \to AD \perp PC$

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@user2612743 $AC$ is the diameter. – hhsaffar Dec 21 '13 at 21:40
Triangle $ADP$ is a right triangle, with the right angle at $D.$ Segment $AP = 18,$ with an intervening point $B$ which divides it so that $AB = 10$ and $BP = 8.$ Segment $PD = 9.$ Segment $DA = 9\sqrt{3}.$ Points $A,$ $B,$ and $D$ determine the required circle.
Now let's locate points on the Cartesian plane. $$A=\left(-\frac{9\sqrt 3}{2},0\right), \ \ D=\left(\frac{9\sqrt 3}{2},0\right), \ \ P=\left(\frac{9\sqrt 3}{2},9\right), \ \ B=\left(\frac{\sqrt 3}{2},5\right).$$ The $y$-axis is the perpendicular-bisector of $AD.$ The intersection of the perpendicular-bisector of either $AB$ or $BD$ with the $y$-axis is the center of the required circle.