Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 6 - Section 6.2 - More Angle Measures in the Circle - Exercises - Page 288: 49

Answer

(a) $\triangle ABE \cong \triangle DFC$ (b) $\angle DAB$ and $\angle BCD$ are supplementary.

Work Step by Step

(a) Since $\angle ABD$ subtends the diameter of the circle, $m\angle ABD = 90^{\circ}$ Thus $m\angle ABE = 90^{\circ}$ It is given that $m\angle DFC = 90^{\circ}$ Thus $\angle DFC \cong \angle ABE$ Since $\angle ACD$ subtends the diameter of the circle, $\angle ACD = 90^{\circ}$ Thus $\angle ECD = 90^{\circ}$ Then $m\angle ECF + m\angle FCD = 90^{\circ}$ Since $m\angle FEC + m\angle ECF = 90^{\circ}$, then $m\angle FEC = m\angle FCD$ Also, $m\angle FEC = m\angle BEA$ since they are opposite angles. Thus $m\angle FCD = m\angle BEA$, and so $\angle FCD \cong \angle BEA$ Since two angles of the triangles $\triangle ABE$ and $\triangle DFC$ are congruent, the third angle must also be congruent since the sum of three angles in a triangle is $180^{\circ}$ Therefore, $\triangle ABE \cong \triangle DFC$ (b) Since the quadrilateral $ABCD$ is a cyclic quadrilateral, the sum of opposite angles in the quadrilateral is $180^{\circ}$ Therefore, $m\angle DAB + m\angle BCD = 180^{\circ}$, so $\angle DAB$ and $\angle BCD$ are supplementary.
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