Question 1152930: abcd is a quadrilateral inscribed in a circle. bc = cd, ab//dc and angle dbc =50°, find angle adb
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! For a quadrilateral to be in a circle with two equal sides and two parallel sides, it has to be a square. If the opposite sides aren't equal, it isn't parallel, and if they are equal, the adjacent sides can't be equal (rectangle that is not a square)
therefore ADB is 50 degrees as well
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
(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
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