The measure of an inscribed angle is one-half the measure of its intercepted arc.
We will start with Case I, where the center of ⊙ Q is on a side of the inscribed angle.
Let's recall the Given and Prove statements!
Given:& ∠ ABC is inscribed in ⊙ Q.
& Let m∠ B=x^(∘).
& Center Q lies on BC.
Prove:& m∠ ABC=1/2mAC
We will continue by showing that △ AQB is an isosceles triangle. Remember that radii of a circle are congruent.
Therefore, AQ and BQ are also congruent.
AQ≅ BQ
By the definition of an isosceles triangle, we can say that △ AQB is an isosceles triangle.
△ AQB is isosceles.
To proceed to the next step, we should recall the Base Angles Theorem (Theorem 5.6).
Base Angles Theorem
If two sides of a triangle are congruent, then the angles opposite them are congruent.
Using this theorem, we can conclude that the measure of ∠ A is x^(∘).
m∠ A = x^(∘)
From here, we will write mAC in terms of x.
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent angles.
By this theorem, the measure of ∠ AQC is 2x^(∘).
m∠ AQC = 2x^(∘)
Looking at the diagram, we can see that ∠ AQB is a central angle, and its circular arc is AC. Since the measure of a circular arc is the measure of its central angle, the measure of AC is also 2x^(∘).
mAC = 2x^(∘)
Now that we know mAC and m∠ ABC in terms of x, we can write an equality between them.
First, we will substitute 12(2x) for x into m∠ ABC=x.
m∠ABC = 1/2(2x)
Finally, by substituting mAC for 2x we can complete our proof.
m∠ABC = 1/2mAC
Let's summarize the above process in a flow proof.
b In this part, we will prove the second case of the theorem. Here, the center Q of the circle is inside the inscribed angle.
Let's first write the Given and Prove statements for Case II.
Given:& ∠ ABC inscribed in ⊙ Q.
& DB is a diameter.
Prove:& m∠ ABC=1/2mAC
Now, we can continue with the Angle Addition Postulate (Postulate (1.4).
Notice that m∠ ABC is the sum of m∠ ABD and m∠ DBC.
m∠ ABC= m∠ ABD+ m∠ DBC
Since the center of the circle is on the sides of two angles, we will next consider the Inscribed Angle Theorem Case I.
From Case I, we know that m∠ ABD is half of mAD and m∠ DBC is half of mDC.
m∠ ABD=1/2mAD
m∠ DBC=1/2mDC
Next, we will substitute 12mAD and 12mDC for m∠ ABD and m∠ DBC into the equation m∠ ABC=m∠ ABD+m∠ DBC.
m∠ ABC=1/2mAD+1/2mDC
Let's factor out 12.
m∠ ABC=1/2(mAD+mDC)
Finally, since the sum of mAD and mDC gives us mAC, we can complete our proof by using the Arc Addition Postulate (Postulate 10.1).
m∠ ABC=1/2 mAC
Let's summarize the above process using a flow proof.
c Finally, we will prove Case III.
As we did in Part B, we will begin by writing the Given and Prove statements for Case III.
Given:& ∠ ABC inscribed in ⊙ Q.
& DB is a diameter.
Prove:& m∠ ABC=1/2mAC
From here, following the exact same steps from Part B we can also prove Case III.
