Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 242: 45

Answer

(a) $\triangle ABE$ ~ $\triangle CFE$ (b) $\triangle CFE$ ~ $\triangle DFC$ (c) $\triangle ABE$ ~ $\triangle DFC$

Work Step by Step

(a) We can find $m\angle CFE$: $m\angle CFE + m\angle CFD = 180^{\circ}$ $m\angle CFE + 90^{\circ} = 180^{\circ}$ $m\angle CFE = 90^{\circ}$ We can find two congruent angles in $\triangle ABE$ and $\triangle CFE$: $\angle ABE \cong \angle CFE$, since both angles are $90^{\circ}$ $\angle BEA \cong \angle FEC$, since these angles are opposite angles Since there are two congruent angles in $\triangle ABE$ and $\triangle CFE$, all three angles must be congruent. Therefore, $\triangle ABE$ ~ $\triangle CFE$ (b) We can find two congruent angles in $\triangle CFE$ and $\triangle DFC$: $\angle CFE \cong \angle DFC$, since both angles are $90^{\circ}$ $\angle FCE \cong \angle CDF$, since $m\angle FCE = 90^{\circ} - m\angle FCD = m\angle CDF$ Since there are two congruent angles in $\triangle CFE$ and $\triangle DFC$, all three angles must be congruent. Therefore, $\triangle CFE$ ~ $\triangle DFC$ (c) $\triangle ABE$ ~ $\triangle CFE$ ~ $\triangle DFC$ Therefore, $\triangle ABE$ ~ $\triangle DFC$
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