INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

October 27, 2013

Sides of a Star

If the points A, B, C, D and E in the figure lie on a circle, then \frac {AD}{BE}   equals ?

660First join AB, BC, CD, DE and EA. Using the fact that angles in the same segment are equal, we mark the newly created angles as in "New Figure".

Finally we use sine rule \frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c} = \frac {1}{2R} = \frac {2 \Delta}{abc} where R is the circumradius.

We use triangle ABD and triangle BEC and apply the sine rule in each of them to get \frac {\sin (B+C) }{AD} = \frac {1}{2R} and \frac {\sin (C+D) }{BE} = \frac {1}{2R}

Hence \frac {AD }{BE} = \frac {\sin(B+C)}{\sin(C+D)}

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