bExample Answer: m∠ A = m∠ D = 30^(∘) and mAC=mBD=60^(∘) Explanation: The arcs are congruent
C
cConjecture: In a circle, two parallel chords cut congruent arcs.
D
d mPR = mQS = 70^(∘)
Practice makes perfect
a Let's begin by drawing a circle using a compass. We will also mark three points on it.
Next, we will draw the chord AB and the segment AD.
To draw the second chord CD parallel to AB, with the same compass setting we draw an arc centered at A and another centered at D.
Next, we set the compass so that both ends align with F and G. With this compass setting, place the compass at H and draw an arc that intersects the arc drawn previously.
Finally, we draw the chord CD which passes through D and the intersection point I.
As we can see, m∠ A = 30^(∘). Similarly, let's find m∠ D.
As before, we have found that m∠ D = 30^(∘). To determine the measure of AC and BD, we use the Inscribed Angle Theorem.
c
mAC = 2m∠ D^(30^(∘))
⇓
mAC = 60^(∘)
c
mBD = 2m∠ A^(30^(∘))
⇓
mBD = 60^(∘)
Since both arcs have the same measure, we conclude that they are congruent.
c Let's draw a second circle and two parallel chords.
Next, we draw JM and find m∠ J with the help of a protractor.
As we can see, m∠ J=40^(∘). Similarly, let's find m∠ M.
Once more, we got that m∠ M=40^(∘). To find the measures of the intercepted arcs we will use the Inscribed Angle Theorem.
c
mKM = 2m∠ J^(40^(∘))
⇓
mKM = 80^(∘)
c
mJL = 2m∠ M^(40^(∘))
⇓
mJL = 80^(∘)
Since both arcs have the same measure, we conclude, as in Part B, that the arcs are congruent. This leads us to set the following conjecture.
In a circle, two parallel chords cut congruent arcs.
d We are given the figure below, where the chords are parallel.
According to the conjecture we wrote in the previous part we know both arcs are congruent, which implies that mQS = mPR.
6x-26 = 4x+6Let's solve the equation above for x.
Substitute x=16 into any of the arcs measures.
mQS = 4(16)+6 = 70^(∘) = mPR
Using Inscribed Angles
To verify that the results we got are correct, let's use the inscribed angles.
Because ∠ R and ∠ Q are alternate interior angles and PR∥ RS, we get that ∠ R ≅ ∠ Q. Thus, they have the same measure.
m ∠ R = m ∠ Q
Then, by applying the Inscribed Angle Theorem we have that the measure of the inscribed angle is one half the measure of the intercepted arc.
m ∠ R = 12mQS = 12(4x+6)
m ∠ Q = 12mPR = 12(6x-26)
Let's solve this system for x.
Now that we know the value of x, let's substitute it into the arc measures.
mQS &= 4(16)+6 = 70^(∘)
mPR &= 6(16)-26 = 70^(∘)
As we can see, both arcs have a measure of 70^(∘), which tells us that our first result is correct.
