Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 3 - Section 3.3 - Isosceles Triangles - Exercises: 20


$\angle$ABC=35$^{\circ}$ $\angle$ADC=35$^{\circ}$ $\angle$1=110$^{\circ}$

Work Step by Step

In concave quadrilateral ABCD $\angle$A=40$^{\circ}$ $\triangle$ABD is isosceles. That means, $\angle$ABD=$\angle$ADB. BC bisects $\angle$ABD which means, $\angle$ABC=$\angle$CBD Also, DC bisects $\angle$ADB which means, $\angle$ADC=$\angle$CDB Lets assume $\angle$ABD=$\angle$ADB=x so $\angle$ABD+$\angle$ADB=2x $\angle$A+$\angle$ABD+$\angle$ADB=180$^{\circ}$ 40$^{\circ}$+2x=180$^{\circ}$ 2x=180$^{\circ}$-40$^{\circ}$ x=$\frac{140}{2}$ x=70$^{\circ}$ $\angle$ABD=$\angle$ADB=70$^{\circ}$ $\angle$ABC=$\angle$CBD=$\angle$ABD$\div$2 $\angle$ABC=$\angle$CBD=35$^{\circ}$ $\angle$ABC=$\angle$ADC=35$^{\circ}$ now $\angle$1+$\angle$CBD+$\angle$CDB=180$^{\circ}$ $\angle$1+35$^{\circ}$+35$^{\circ}$=180$^{\circ}$ $\angle$1=180$^{\circ}$-70$^{\circ}$ $\angle$1=110$^{\circ}$
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