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June 18th, 2016, 02:07 PM   #1
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Find the ratio of lines in a circle

I have no idea how to solve this problem.
ABCD is just an irregular Quadrilateral so nothing too special with that figure.
We are looking for the ratio BC:CD and we only have that two angles. I know that the answer is 1:√2, but I have no idea how to find it.
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June 18th, 2016, 03:29 PM   #2
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Quote:
Originally Posted by mitaka90 View Post
I have no idea how to solve this problem.
ABCD is just an irregular Quadrilateral so nothing too special with that figure.
We are looking for the ratio BC:CD and we only have that two angles. I know that the answer is 1:√2, but I have no idea how to find it.
There is plenty special with the figure. It is a cyclic quadrilateral. That means that $\displaystyle \angle B$ and $\displaystyle \angle D$ are supplementary. So their sines must be equal.

With that in mind, use the law of sines on $\displaystyle \triangle ABC$ and $\displaystyle \triangle ACD$.
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June 18th, 2016, 05:50 PM   #3
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I meant that there isn't something special like trapezoid or any other specific figure and I'm pretty sure it's more than obvious that it is a cyclic quadrilateral. Ty, but I was already helped from a person from another site, who was actually faster and almost solved the problem for me, thus easier to understand. No offence, tho.
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June 18th, 2016, 07:36 PM   #4
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By the inscribed angle theorem, $\angle{CDB}=30^\circ$ and $\angle{CBD}=45^\circ$.

Construct the altitude of $\triangle{BCD}$ from $C$ to $\overline{BD}$ and, without a loss of generality, assign it a length of $1$ unit. Now it's easy to determine the (resulting) lengths of $\overline{BC}$ and $\overline{CD}$ and to compute the required ratio.
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June 19th, 2016, 11:37 AM   #5
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Oh, so you're here as well, greg? It seems you're being helpful everywhere. Thank you for the alternative solution, but I liked Serena's more, since it was easier to understand and used stuff that I learned this year. I've forgotten a lot of the arcs properties and generally the feel for solving problems with them.
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