We are asked to find two possible measures of the arc AE. To do so, we will draw two diagrams that could represent the situation described in the exercise. Let's start by drawing the circle ⊙ R and labeling two points on the circle, A and B, so that mAB=60^(∘).
Now we want to place C so that mBC=25^(∘). Note that we can place C on either side of B.
Similarly, we can place D on either side of C, and place E on either side of D. All diagrams obtained in this way, could represent the situation. Since we are only looking for two possible measures of AE, we will consider two of these diagrams.
Diagram I
In the first diagram we assume that C is on the left-hand side of B.
We will also place D on the left-hand side of C, and E on the left-hand side of D. Remember that mCD= 70^(∘) and mDE= 20^(∘).
Finally, let's determine the measure of the minor arc AE.
We can see that AE consists of AB, BC, CD, and DE. Note that the Arc Addition Postulate is true for any number of adjacent arcs. Therefore, the measure of AE is equal to the sum of measures of AB, BC, CD, and DE.
mAE=mAB+mBC+mCD+mDE
We know that mAB= 60^(∘), mBC= 25^(∘), mCD= 70^(∘), and mDE= 20^(∘). Let's substitute these values into the above expression.
In the second diagram, we assume that C is on the right-hand side of B.
We will place D on the left-hand side of C, and E on the left-hand side of D. Remember that mCD= 70^(∘) and mDE= 20^(∘).
Now, let's determine the measure of AE in this case.
We can see that AE consists of AB, BD, and DE. We will use the Arc Addition Postulate once more.
mAE=mAB+mBD+mDE
We know that mAB= 60^(∘) and mDE= 20^(∘). To find mAE, we have to find mBD first.
Note that BC and BD form CD, so we can write the following.
mCD=mBC+mBD
Let's substitute mCD= 70^(∘) and mBC= 25^(∘) into the above equation and solve it for mBD.
