Core Connections Integrated II, 2015
CC
Core Connections Integrated II, 2015 View details
Chapter Closure

Exercise 119 Page 593

a We know that AC is the diameter of ⊙ E. Since m∠ AEB is a right angle, it must be that m∠ BEC is also a right angle because these angles form a linear pair. This also means that △ BEC is an isosceles right triangle with base angles of 45^(∘). Let's add this information to the diagram.
a circle with a diameter AC and two radii

Notice that ∠ BCE and ∠ CED are alternate interior angles. Since BC and ED are parallel segments, we know by the Alternate Interior Angles Theorem that these angles are congruent.

a circle with a diameter AC and two radii

We see that CD is the intercepted arc of the central angle ∠ CED. Therefore, it must be that mCD is also 45^(∘).

b From the diagram we see that BC is 112^(∘). We also see that BC is the intercepted arc to the inscribed angle ∠ BAC. According to the Inscribed Angle Theorem, the inscribed angle is always half the measure of its intercepted arc. Therefore, ∠ BAC must be 56^(∘).
a circle with an inscribed triangle

Now we can find the measure of ∠ ACB by using the Interior Angles Theorem. 56^(∘)+65^(∘)+m∠ C=180^(∘) ⇓ m∠ C=59^(∘) Notice that ∠ ACB is the inscribed angle to the intercepted arc AB. By using the Inscribed Angles Theorem again, we can determine AB to be double that of ∠ ACB. That is, 2(59^(∘)) =118^(∘).

a circle with an inscribed triangle