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June 18th, 2016, 02:07 PM  #1 
Newbie Joined: Jun 2016 From: Bulgaria Posts: 3 Thanks: 0  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. 
June 18th, 2016, 03:29 PM  #2  
Senior Member Joined: Feb 2010 Posts: 673 Thanks: 125  Quote:
With that in mind, use the law of sines on $\displaystyle \triangle ABC$ and $\displaystyle \triangle ACD$.  
June 18th, 2016, 05:50 PM  #3 
Newbie Joined: Jun 2016 From: Bulgaria Posts: 3 Thanks: 0 
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.

June 18th, 2016, 07:36 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,806 Thanks: 1045 Math Focus: Elementary mathematics and beyond 
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. 
June 19th, 2016, 11:37 AM  #5 
Newbie Joined: Jun 2016 From: Bulgaria Posts: 3 Thanks: 0 
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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circle, find, lines, ratio 
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