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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 6.CR, Problem 9CR
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Elementary Geometry for College St...

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

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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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In Review Exercises 5 to 10, B A → is tangent to the circle at point A in the figure shown.

G i v e n :           C   i s   t h e   m i d p o i n t   o f   A C D ^   a n d     m ∠ B = 40 °

      F i n d :             m A D ^ ,   m A C ^ , m D C ^

Chapter 6.CR, Problem 9CR, In Review Exercises 5 to 10, BA is tangent to the circle at point A in the figure shown.

To determine

To find: mAD^,mAC^, mDC^

Explanation

Theorem:

1) The measure of an angle formed when two secants intersect at a point outside the circle is one-half the difference of the measures of two intercepted arcs.

2) The measure of an inscribed angle of a circle is one-half the measure of its intercepted arc.

Calculation:

Given that

m∠B=40° and C is the midpoint of mACD^ and BA→ is tangent to the circle at point A.

By theorem, the measure of an angle (m∠B) formed when two secants (BD and BA) intersect at a point (B) outside the circle is one-half the difference of the measures of two intercepted arcs AD^ and AC^ .

So, m∠B=12(mAD^-mAC^)

40°=12(mAD^-mAC^)

2×40°=mAD^-mAC^

mAD^-mAC^=80°

So, mAD^=mAC^+80°

Now, mAD^+mAC^+mDC^=360°

Since, C is the midpoint of mACD^, mAC^=mDC^

mAD^+

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