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Chapter 16: Chapter 16

Problem 298

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Prove that inscribed angles which intercept the same arc are congruent.

Short Answer

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By drawing a circle with center O and two inscribed angles ∠ACB and ∠ADB intercepting the same arc AB, we can form two triangles: triangle ADC and triangle BDC. Applying the inscribed angle theorem, we find that m(∠ACB) = \(\frac{1}{2}\)m(arc AB) and m(∠ADB) = \(\frac{1}{2}\)m(arc AB). Since both inscribed angles are equal to half the measure of the intercepted arc AB, we can conclude that m(∠ACB) = m(∠ADB), proving that inscribed angles which intercept the same arc are congruent.
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Draw a circle with two inscribed angles intercepting the same arc

Draw a circle centered at O. Let A, B, and C be three points on the circle forming the arc AB. Now, draw two inscribed angles ∠ACB and ∠ADB that intercept the same arc AB. The points A and B are the endpoints of the intercepted arc, and C and D are the vertices of the inscribed angles.

Step 2: Construct the chord connecting points C and D

Draw the chord CD connecting points C and D. This forms two triangles: triangle ADC and triangle BDC.

Step 3: Calculate the measure of the angle at the center of the circle

Since A and B are the endpoints of the intercepted arc, the angle ∠AOB at the center of the circle is called the central angle. The measure of this angle is equal to the measure of the intercepted arc AB, denoted as m(arc AB).

Step 4: Calculate the measure of the angles in the triangles ADC and BDC

Use the property that in a triangle, the sum of the measures of the internal angles is 180 degrees. In triangle ADC, we have: m(∠ADC) + m(∠DAC) + m(∠ACD) = 180 degrees In triangle BDC, we have: m(∠BDC) + m(∠DBC) + m(∠BCD) = 180 degrees

Step 5: Use circle properties to find the relationship between the central angle and inscribed angles

By the inscribed angle theorem, the measure of an inscribed angle is half the measure of the intercepted arc. So, we have: m(∠ACB) = \(\frac{1}{2}\)m(arc AB) m(∠ADB) = \(\frac{1}{2}\)m(arc AB)

Step 6: Substitute the relationship in the triangles ADC and BDC

Substitute the relationship between inscribed angles and intercepted arc in the equations for triangles ADC and BDC: For triangle ADC: m(∠ADC) + m(∠DAC) + 2m(∠ACB) = 180 degrees For triangle BDC: m(∠BDC) + m(∠DBC) + 2m(∠ADB) = 180 degrees

Step 7: Compare the measures of the inscribed angles

Since both m(∠ACB) and m(∠ADB) are equal to half the measure of the intercepted arc AB, it follows that: m(∠ACB) = m(∠ADB) Thus, we have proved that inscribed angles which intercept the same arc are congruent.

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Most popular questions from this chapter

Chapter 16

Given: Points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\) are in \(\mathrm{OP} ; \mathrm{AB} \cong \mathrm{AD}^{\prime}\); \(\mathrm{BC}^{-} \cong \mathrm{DC}^{-}\) Prove: $\angle \mathrm{B} \cong \angle \mathrm{D}$
Chapter 16

Let \(\angle \mathrm{B}\) be an angle inscribed in a circle and let it have measure greater than \(90^{\circ}\). (See figure.) Prove that $$ \mathrm{m} \angle \mathrm{B}=180^{\circ}-(1 / 2) \mathrm{m} \angle \mathrm{P} $$
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Let \(\mathrm{m} \angle \mathrm{A}=90^{\circ}\) in \(\triangle \mathrm{ABC}\). Let \(\mathrm{D}, \mathrm{E}\), and \(\mathrm{F}\) be the midpoints of $\underline{A B}, \underline{A C}\(, and \)\underline{B C}\(, respectively. Prove that \)F$ is the center of a semicircle which contains \(\mathrm{B}, \mathrm{A}\) and \(\mathrm{C}\).
Chapter 16

Given two intersecting chords of a circle, show that the measure of the angle formed by the intersection is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle.
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Let \(\angle \mathrm{A}\) be inscribed in a circle, and let $\mathrm{m} \angle \mathrm{A}<90^{\circ} .\( Let \)\angle \mathrm{P}$ be the angle, with vertex at the center of the circle, which intercepts the same arc as $\angle \mathrm{A}\(. (Then \)\angle \mathrm{A}\( and \)\angle \mathrm{P}$ are said to be related). Prove: $\mathrm{m} \angle \mathrm{A}=(1 / 2) \mathrm{m} \angle \mathrm{P}$.
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