Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 6 - Section 6.3 - Line and Segment Relationships in the Circle - Exercises - Page 306: 50

Answer

Since $AD = CD$ and $AB = BC$, the quadrilateral $ABCD$ is a kite.

Work Step by Step

We know that $\angle DAB = \angle DCB = 90^{\circ}$ $AD = CD$, because each line is a radius of the circle. We can use the Pythagorean theorem to show that the length of $AB$ is equal to the length of $BC$: $AB = \sqrt{(BD)^2-(AD)^2}$ $AB = \sqrt{(BD)^2-(CD)^2}$ $AB = BC$ Since $AD = CD$ and $AB = BC$, the quadrilateral $ABCD$ is a kite.
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