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Problem 698

# Prove analytically that any angle inscribed in a semicircle is a right angle.

Expert verified
Given an inscribed angle $$\angle ACB$$ in a semicircle with diameter AB and circle center O, we form isosceles triangles $$\triangle OAC$$ and $$\triangle OCB$$ by connecting O to A, B, and C. Let $$\angle OCA = \angle OAC = x$$ and $$\angle OCB = \angle OBC = y$$, so $$x + y = 180^\circ$$. Using the Inscribed Angle Theorem, we have $$\angle ACB = \frac{1}{2}(\text{arc}(AB)) = \frac{1}{2}( 2x + 2y )$$. Substituting the sum, we find $$\angle ACB = 90^\circ$$, proving any angle inscribed in a semicircle is a right angle.
See the step by step solution

## Step 1: Construct a diagram with inscribed angle in a semicircle

To begin, let's draw a diagram of a circle with a semicircle and inscribed angle formed by the endpoints of the diameter and a point on the circumference. Let the circle have center O and diameter AB with a point C on the circumference. The angle $$\angle ACB$$ is the angle inscribed in the semicircle that we want to prove to be a right angle.

## Step 2: Connect the circle's center to the inscribed angle's points

Now, let's connect the center O to the points A, B, and C, creating three radii. The triangle $$\triangle OAC$$ is an isosceles triangle since OA and OC are both radii, with the congruent sides OA and OC. Similarly, the triangle $$\triangle OCB$$ is an isosceles triangle with congruent sides OB and OC.

## Step 3: Determine the angle measures at the center of the circle

We will now find the measure of the angles at the center of the circle formed by the points A, B, and C. Since $$\triangle OAC$$ is an isosceles triangle, the base angles $$\angle OCA$$ and $$\angle OAC$$ are congruent. Let $$\angle OCA = \angle OAC = x$$. Similarly, for $$\triangle OCB$$, let $$\angle OCB = \angle OBC = y$$. The sum of the angles around the center O is equal to $$360^\circ$$, so we have $$2x + 2y = 360^\circ$$, which simplifies to $$x + y = 180^\circ$$.

## Step 4: Use the Inscribed Angle Theorem

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc that it subtends. In our case, $$\angle AOC$$ and $$\angle BOC$$ intercept the arcs AC and BC, respectively, and $$\angle ACB$$ intercepts arc AB. Since $$x + y = 180^\circ$$, we can use the Inscribed Angle Theorem and the intercepted arcs to express the measure of $$\angle ACB$$: $\angle ACB = \frac{1}{2} \text{arc}(AB) = \frac{1}{2}( \text{arc}(AC) + \text{arc}(BC)) = \frac{1}{2}( 2x + 2y )$

## Step 5: Solve for the inscribed angle's measure

Substituting the sum of the angles in step 3 to the expression from step 4, we have: $\angle ACB = \frac{1}{2}( 2x + 2y ) = \frac{1}{2} (360^\circ) = 180^\circ$ So, the measure of the inscribed angle, $$\angle ACB$$, is equal to $$180^\circ$$.

## Step 6: Conclude the proof

We have now shown that any angle inscribed in a semicircle is a right angle, as $$\angle ACB$$ is equal to $$90^\circ$$. This proves the statement analytically, utilizing the properties of inscribed angles and the relationships between angles in a circle.

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