Elementary Geometry for College Students (6th Edition)

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

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.