Elementary Geometry for College Students (6th Edition)

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

Chapter 7 - Section 7.2 - Concurrence of Lines - Exercises - Page 326: 44

Answer

$AF = s-a$

Work Step by Step

We can write an equation for each side of $\triangle ABC$: $BD+CD = a$ $AE+CE = b$ $AF+BF = c$ Then: $BD+CD + AE+CE + AF+BF = a+b+c$ $BD+CD + AF+CD + AF+BD = a+b+c$ $2BD+2CD + 2AF = a+b+c$ $BD+CD + AF = \frac{1}{2}(a+b+c)$ $BD+CD + AF = s$ $a + AF = s$ $AF = s-a$
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