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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

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BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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Note: Exercises preceded by an asterisk are of a more challenging nature.

G i v e n : O w i t h O E ¯ C D ¯ C D = O C F i n d : m C F

Chapter 6.3, Problem 1E, Note: Exercises preceded by an asterisk are of a more challenging nature.

To determine

To find:

To find mCF^.

Explanation

Given that O with OE¯CD¯ and CD=OC

Consider the following figure,

From the above figure shows that OC=OD. Since radius of O is same for all points on circle.

From the given condition CD=OC=OD

Therefore, ΔOCD is equilateral triangle. Therefore, C=COD=D=60

Let us take ΔOCE,ΔOED

Also, OEC=OED=90. Since OE¯CD¯.

Theorem:

If a line is drawn through center of a circle perpendicular to a chord then it bisects the chord and its arc

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