Inscribed Angle Theorem
Our old buddies, central angles, always had their vertices at the center. But an inscribed angle's vertex can be anywhere on the circle. Is the center of the circle even relevant anymore?
It turns out the center of the circle is still super relevant to inscribed angles. The center can help us get organized here. Let's classify all possible central angles ∠ABC based on whether the center O of the circle is "inside," "outside," or "on" the angle, and examine these brief cases.
O is on one of the chords which make up ∠ABC. In other words, one of the sides of ∠ABC is a diameter of ⊙O.
Let x stand for the measure of the inscribed angle ∠ABC, and we'll see if we can find mAC. We'll draw in line segment OA, because if we can find the measure of the central angle ∠AOC, we'll know mAC.
By drawing in OA, we've made ΔAOB, which must be an isosceles triangle since OB and OA are radii of the same circle and are therefore congruent. That means m∠BAO = m∠ABC = x. We've also conveniently made ∠AOC an exterior angle to the triangle, which means that m∠AOC = m∠BAO + m∠ABC = x + x = 2x. So mAC = 2x.
Now we know that a Case 1 angle intercepts an arc with twice the measure of the angle. We can use that knowledge to examine Case 2.
O is in the interior of ∠ABC.
Let's add a diameter in there.
Now we have two Case 1 angles ∠ABD and ∠DBC. From our work on Case 1, we can say that m∠ABD = ½ × mAD and m∠DBC = ½ × mDC
Since ∠ABD and ∠DBC are adjacent angles, m∠ABC = m∠ABD + m∠DBC. Substituting, we have m∠ABC = ½ × mAD + ½ × mDC. Factoring out the common ½, we have m∠ABC = ½ × (mAD + mDC).
Now we bring in our Arc Addition Postulate. Seemed silly at the time, didn't it? Well, it's here to save us now.
mAD + mDC = mAC
Now we substitute that in, and we have m∠ABC = ½ × mAC.
In other words, a Case 2 inscribed angle intercepts an arc with twice the measure of the angle, just as a Case 1 angle does. Curious. Now let's look at Case 3.
O is in the exterior of ∠ABC.
Diameters have been good to us in the past, so how about we throw one in there, just to see what happens? After all, what's the worst that could happen? It's just math.
We can actually use the same strategy we used in Case 2 with a little twist. Here, we have two Case 1 angles: ∠CBD, which intercepts arc CD, and ∠ABD, which intercepts arc AD. From our work in Case 1, we know that m∠ABD = ½ × mAD and m∠CBD = ½ × mCD.
We'll find our target angle by subtraction this time, not by addition (although you can't have subtraction without addition, so in a way, we're still using addition).
If we subtract, we get m∠ABC = m∠CBD – m∠ABD. Substituting, we have m∠ABC = ½ × mCD – ½ × mAD. Factoring, we have m∠ABC = ½ × (mCD – mAD).
Now we bring in our old friend, the Arc Addition Postulate, and we get m∠ABC = ½ × mAC.
We've shown that a Case 3 inscribed angle intercepts an arc with twice the measure of the angle—same as a Case 1 angle or a Case 2 angle. Since any inscribed angle falls into one of the three cases, we've proven the Inscribed Angle Theorem: for ∠ABC inscribed in a circle containing points A and C, mAC = 2 × m∠ABC. Translation: the arc is twice the angle.
Given that ∠ACB is inscribed in a circle containing points A and B, and that m∠ACB = 42°, what is the measure of arc AB?
Here we have an inscribed angle intercepting an arc, so we can use the Inscribed Angle Theorem. We can start out with mAB = 2 × m∠ACB, and then substitute 42° for m∠ACB. That'll give us mAB = 2 × 42°, or mAB = 84°.
Given that ∠ACB is inscribed in circle ⊙O with radius 5 cm containing points A and B, and that m∠ACB = 78°, what is the length of arc AB?
Before we calculate arc length, we can calculate the measure of the arc using the Inscribed Angle Theorem:
mAB = 2 × m∠ACB = 2 × 78° = 156°
Now we can use the arc length formula , with θ = 156° and r = 5 cm. This should give us an arc length of about 13.6 cm.