Multiple choice

$AB$ is a diameter of the circle, $CD$ is the chord equal to the radius the of the circle. $AC$ and $BD$ when extended intersect at a point $E$. Then find measure of $\angle AEB$.

  1. ${30^ \circ }$
  2. ${60^ \circ }$
  3. ${90^ \circ }$
  4. ${120^ \circ }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

AB is diameter. CD = radius. Triangle OCD is equilateral (OC=OD=CD=r). Angle COD = 60 degrees. Angle CBD = 1/2 angle COD = 30 degrees (angle at circumference). Angle ACB = 90 degrees (angle in semicircle). In triangle BCE, angle CBE = 30, angle BCE = 180 - 90 = 90. Angle CEB = 180 - 90 - 30 = 60 degrees.

AI explanation

In triangle OBC, sides OC and OB are radii, and chord CD is given as equal to the radius, making it an equilateral triangle with an angle of 60 degrees at the center. Since AB is the diameter, angle BAC is an angle in a semicircle, so angle BAC = 90 degrees. The angle CAD is found by subtracting the central angle from 90 degrees, giving angle CAD = 30 degrees. In triangle CAE, using the exterior angle theorem where angle CAB is an exterior angle for the triangle formed at E, angle AEB equals 90 degrees minus 30 degrees, which results in 60 degrees.

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