Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 6 - Review Exercises - Page 307: 27

Answer

First of all, we know that angle DEB is congruent to angle AEC by the vertical angles theorem. In addition, since the radius of a circle is always perpendicular to the tangent line at that point, it follows that BDE is congruent to ACE. Thus, by AA, the triangles are similar, allows us to set up this proportion: $\frac{BD}{AC} = \frac{ED}{CE} \\ BDCE = ACED$
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