Pregunta
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The figure below shows a circle with center \( E \) and diameter \( \overline{X G} \). Which of the angles must be right angles? Select all that apply.

Ask by Wright Pope. in the United States
Mar 11,2025

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All inscribed angles that intercept the diameter \( \overline{XG} \) are right angles.

Solución

Since \( \overline{XG} \) is a diameter, any inscribed angle that intercepts the arc \( \widehat{XG} \) must be a right angle. (This is a well‐known fact often called the “angle in a semicircle theorem”.) In other words, if a point \( P \) (other than \( X \) or \( G \)) lies on the circle, then the inscribed angle \( \angle XPG \) intercepts the semicircular arc \( \widehat{XG} \) and must satisfy \[ \angle XPG=90^\circ. \] Thus, any angle in the circle whose sides pass through \( X \) and \( G \) (and whose vertex is on the circle) must be a right angle. So the answer is: All inscribed angles that intercept the diameter \( \overline{XG} \) are right angles.

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Extra Insights

In a circle, any angle inscribed with its vertex on the circle and whose sides are formed by the endpoints of a diameter is a right angle. Therefore, if the diameter is \( \overline{X G} \) and \( E \) is the center, any angle formed by \( X \) or \( G \) and any other point on the circle (let's say \( P \)) must be a right angle (i.e., \( \angle X P G \) is a right angle). Additionally, if you have triangles formed with the diameter as one side and the other vertices on the circumference, those angles opposite the diameter will also always be right angles due to the inscribed angle theorem. So, if you see angles directly related to that diameter, you should consider them as right angles!

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