In triangle ABE, A and B lie on a circle (centered at O), and E lies outside the circle. BE intersects the circle at C, AE intersects at D. (Thus, quadrilateral ABCD is inscribed in the circle.) AB:BC:CD:DA are in the ratios 10:7:7:12. What is the length of BE?
This is from ARCO's Preparation for the Praxis II Exam 2005, and I'm really interested in how to solve this more than the answer. (The question, #17 on p.155, actually gives AB's length as 8 and offers 4 nearest-integer answers: 50, 37, 31, 29.) Decades ago, I'd have aced this exam without prep, but I need to refresh what I need to know.
So "remember that ..." hints would be perfect for me. Thanks.
Trigonometry problem - pointer wanted?
It is fairly easy to prove that triangles EDC and EAB are similar, i.e. have all their angles equal. This might help because you can then use ratios of side lengths.
Reply:There is a theorem dealing with secants (BE and AE are both secants). BC * CE = AD * DE
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