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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:

enter image description here

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
up vote 4 down vote accepted


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

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How does that help you? From this You can find AD with pythagorean theorem, but how does that help you find the radius? – 1110101001 Dec 21 '13 at 21:38
@user2612743 $AC$ is the diameter. – hhsaffar Dec 21 '13 at 21:40
@user2612743 The triangle ADC is right-angled at D and AC is a diameter, hence you can apply the pythagorean theorem. – RicardoCruz 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.

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