AMC 8 Daily Practice - Circles

Complete problem set with solutions and individual problem pages

Problem 6 Easy

As shown in the figure, quadrilateral ABCD is inscribed in circle O, AB = CD, A is the midpoint of arc BD, and \angle BDC = 60^\circ. What is the measure of \angle ADB?

  • A.

    30

  • B.

    40^\circ

  • C.

    60^\circ

  • D.

    65^\circ

  • E.

    80^\circ

Answer:B

Connect AO, BO, CO, and DO.

Since AB = CD, the arcs they subtend are equal, so arc AB = \text{arc } CD.

Because A is the midpoint of arc BD, arc AB = \text{arc } AD.

Therefore, arc AB = \text{arc } AD = \text{arc } CD.

\angle BDC is a circumferential angle subtended by arc BC, and \angle BDC = 60^\circ.

The central angle subtended by the same arc is twice the circumferential angle, so \angle BOC = 2 \times 60^\circ = 120^\circ.

The total degree measure of the circle is 360^\circ, so the sum of the central angles subtended by arcs AB, AD, and CD is: 360^\circ - \angle BOC = 360^\circ - 120^\circ = 240^\circ.

Since arc AB = \text{arc } AD = \text{arc } CD, their corresponding central angles are equal: \angle AOB = \angle AOD = \angle COD = \frac{240^\circ}{3} = 80^\circ.

\angle ADB is a circumferential angle subtended by arc AB , so it is half of the central angle \angle AOB: \angle ADB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 80^\circ = 40^\circ.

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