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McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Secants, Tangents, and Angle Measures
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Exercise 30 Page 765

Use the Inscribed Angle Theorem and the Triangle Exterior Angle Theorem.

Statements
Reasons
1.
Secants AD and AE
1.
Given
2.
m∠ ECD = 1/2mDE [0.15cm] m∠ BDC = 1/2mBC
2.
Inscribed Angle Theorem
3.
m∠ ECD = m∠ BDC + m∠ A
3.
Triangle Exterior Angle Theorem
4.
1/2mDE = 1/2mBC + m∠ A
4.
Substitution
5.
m∠ A = 1/2mDE - 1/2mBC
5.
Solving for m∠ A
6.
m∠ A = 1/2(mDE - mBC)
6.
Factor out 12
Practice makes perfect

Let's consider a circle and two secants AD and AE.

Circle with two secants from the same exterior point
By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc. m∠ ECD = 12m DE & (I) m∠ BDC = 12m BC & (II) Notice that ∠ ECD is an exterior angle of △ ACD. Then, applying the Triangle Exterior Angle Theorem, we conclude that its measure is equal to the sum of the measures of the two nonadjacent interior angles.

m∠ ECD = m∠ BDC + m∠ A ⇓ m∠ A = m∠ ECD - m∠ BDC Finally, we substitute the two equations found above into the last equation. That way, we get the desired equation. m∠ A = 1/2m DE - 1/2m BC ⇓ m∠ A = 1/2(m DE - m BC)

Two-Column Proof

Given: & SecantsADandAE Prove: & m∠ A = 12(mDE - mBC) Let's summarize the proof we did above in the following two-column proof table.

Statements
Reasons
1.
Secants AD and AE
1.
Given
2.
m∠ ECD = 1/2mDE [0.15cm] m∠ BDC = 1/2mBC
2.
Inscribed Angle Theorem
3.
m∠ ECD = m∠ BDC + m∠ A
3.
Triangle Exterior Angle Theorem
4.
1/2mDE = 1/2mBC + m∠ A
4.
Substitution
5.
m∠ A = 1/2mDE - 1/2mBC
5.
Solving for m∠ A
6.
m∠ A = 1/2(mDE - mBC)
6.
Factor out 12
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Exercises 2 (Page 763)
Exercises 3 (Page 763)
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