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

First 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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