SOLUTION: ABCD is a quadrilateral inscribed in a circle. BC=CD, AB//DC and angle DBC= 50° find angle ADB.

Question 1152941: ABCD is a quadrilateral inscribed in a circle.
BC=CD, AB//DC and angle DBC= 50° find angle ADB.
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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(1)   The given quadrilateral is an ISOSCELES TRAPEZOID.


(2)  The angles at the bases of an isosceles trapezoid are congruent.


(3)  Therefore, angle BCD is congruent to the angle ADC.


(4)  The triangle BDC is isosceles; so its angles DBC and BDC are congruent and have a measure of 50° each.


     Hence, the measure of the angle BCD is 180° - 50° - 50° = 80°.


(5)  Since the angle ADC is congruent to angle BCD, the measure of the angle ADC is 80°.


(6)  So, we have now that the measure of the angle ADC is 80°, and the measure of the angle BDC is 50°.


     Hence, the measure of the angle ADB is 80° - 50° = 30°.    ANSWER

S O L V E D.

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