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.

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.

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^(∘).

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^(∘).

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Exercise 119 Page 593

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