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


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: $\frac{BD}{AC} = \frac{ED}{CE} \\ BDCE = ACED$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.