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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 31 Page 765

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

Statements
Reasons
1.
Tangent FM and secant FL
1.
Given
2.
m∠ MHL = 1/2mLH m∠ GLH = 1/2mGH
2.
Inscribed Angle Theorem
3.
m∠ MHL = m∠ GLH + m∠ F
3.
Triangle Exterior Angle Theorem
4.
1/2mLH = 1/2mGH + m∠ F
4.
Substitution
5.
m∠ F = 1/2mLH - 1/2mGH
5.
Solving for m∠ F
6.
m∠ F = 1/2(mLH - mGH)
6.
Factor out 12
Practice makes perfect

Let's consider a circle, a tangent FM, and a secant FL.

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∠ MHL = 12m LH & (I) m∠ GLH = 12m GH & (II) Notice that ∠ MHL is an exterior angle of △ FHL. 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∠ MHL = m∠ GLH + m∠ F ⇓ m∠ F = m∠ MHL - m∠ GLH Finally, we substitute the two equations found above into the last equation. That way we get the desired equation. m∠ F = 1/2m LH - 1/2m GH ⇓ m∠ F = 1/2(m LH - m GH)

Two-Column Proof

Given: & TangentFMand secantFL Prove: & m∠ F = 12(mLH - mGH) Let's summarize the proof we did above in the following two-column proof table.

Statements
Reasons
1.
Tangent FM and secant FL
1.
Given
2.
m∠ MHL = 1/2mLH m∠ GLH = 1/2mGH
2.
Inscribed Angle Theorem
3.
m∠ MHL = m∠ GLH + m∠ F
3.
Triangle Exterior Angle Theorem
4.
1/2mLH = 1/2mGH + m∠ F
4.
Substitution
5.
m∠ F = 1/2mLH - 1/2mGH
5.
Solving for m∠ F
6.
m∠ F = 1/2(mLH - mGH)
6.
Factor out 12
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arrow_right Exercises p.763-767
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Exercises 2 (Page 763)
Exercises 3 (Page 763)
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