Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 3 - Section 3.2 - Corresponding Parts of Congruent Triangles - Exercises - Page 152: 42

Answer

The statement is true.

Work Step by Step

In a regular pentagon all interior angles equal to $108^{\circ}$. $ABE$ and $BCD$ are isosceles triangles, as they have two equal angles (as given in the HINT). If $\angle EAB = 108^{\circ}$, then $\angle ABE= \frac{180^{\circ}-108^{\circ}}{2}= 36^{\circ}$. With the same reasoning $\angle CBD=36^{\circ}$. Now we need to find $\angle EBD$. We know, that $\angle ABE + \angle CBD + \angle EBD = 108^{\circ}$. So $36^{\circ}+36^{\circ}+ \angle EBD =108^{\circ}$ Therefore: $\angle EBD = 108^{\circ}-36^{\circ}-36^{\circ}=36^{\circ}$. $\angle ABE = \angle CBD =\angle EBD = 36^{\circ}$. So the statement is true.
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.