Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Review Exercises - Page 262: 17

Answer

The proof is in the following form: Column 1 of proof; Column 2 of proof. Thus, we have: 1. ABCD is a parallelogram, and DB intersects AE at point F; given 2. Angle ADC is congruent to angle CBA; opposite angles of parallelograms are congruent. 3. EDF is congruent to ABF; from (2) and the fact that DB cuts the two angles the same way on both sides, it follows that the two are congruent. 4. AFB is congruent to DFE; vertical angles theorem 5. DFE is similar to AFB; AA similarity theorem 6. $\frac{AF}{EF} = \frac{AB}{DE}$; CPSSTP
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