Given that ∠ ADB is a circumscribed angle and ∠ ACB is a central angle whose intercepted arc is AB, we will prove that m∠ ADB=180^(∘)-m∠ ACB.
Given:& ∠ ADB is a circumscribed angle.
&∠ ACB is a central angle.
Prove:& m∠ ADB=180^(∘)-m∠ ACB
Since the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16), let's recall this theorem to start our proof.
Angles Outside the Circle Theorem
If a tangent and a secant, two tangents, or two secants intersect outside a circle, then the measure of the angle formed is one-half the difference of the measures of the intercepted arcs.
As we can see in the figure, there are two tangents intersecting outside ⊙ C.
Therefore, by the second case of this theorem we can express the measure of ∠ ADB in terms of the difference of the measures of the intercepted arcs. Note that because the measure of a whole circle is 360^(∘), the measures of the intercepted arcs are mAB and 360^(∘)-mAB.
m∠ ADB = 1/2(360^(∘)-mAB- mAB)
Now we will perform the subtraction.
m∠ ADB = 1/2(360^(∘)-2mAB)
Next, by the Distributive Property we will distribute 12 over the terms within the parentheses.
m∠ ADB = 180^(∘)-mAB
Remember that the measure of a central angle is the measure of its intercepted arc.
m∠ ACB = mAB
Finally, by the Substitution Property of Equality we will substitute m∠ ACB for mAB into the previous equation and complete our proof.
m∠ ADB = 180^(∘)-m∠ ACB
Let's summarize the above process in a flow proof.
