## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 6 - Review Exercises - Page 307: 34

#### Answer

EB = 6 FC = 7 AD = 3

#### Work Step by Step

We know that AD and AF are equal, so we know: $10 - FC = 9 - BD$ We also know that EC equals FC, so: $10 - AF = 13 - BE$ Finally, we know that BD equals BE. From these, it follows that EB must be 3 more than AF. Thus, we call AF the variable y to obtain: $EB = 3 + y$ $BD = EB = 9 - y$ We solve for y: $3+y = 9 -y \\ y = 3$ Thus: $EB = 3 + 3 = 6$ $AD = AF = y = 3$ $FC = 10 - y = 10 -3 = 7$

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