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Problem 298

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

Expert verified

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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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 $$ \mathrm{m} \angle \mathrm{B}=180^{\circ}-(1 / 2) \mathrm{m} \angle \mathrm{P} $$

Chapter 16

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.

Chapter 16

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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