document.write( "Question 1152941: ABCD is a quadrilateral inscribed in a circle.
\n" ); document.write( "BC=CD, AB//DC and angle DBC= 50° find angle ADB. \n" ); document.write( "

Algebra.Com's Answer #775069 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "(1)   The given quadrilateral is an ISOSCELES TRAPEZOID.\r\n" );
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document.write( "(2)  The angles at the bases of an isosceles trapezoid are congruent.\r\n" );
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document.write( "(3)  Therefore, angle BCD is congruent to the angle ADC.\r\n" );
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document.write( "(4)  The triangle BDC is isosceles; so its angles DBC and BDC are congruent and have a measure of 50° each.\r\n" );
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document.write( "     Hence, the measure of the angle BCD is 180° - 50° - 50° = 80°.\r\n" );
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document.write( "(5)  Since the angle ADC is congruent to angle BCD, the measure of the angle ADC is 80°.\r\n" );
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document.write( "(6)  So, we have now that the measure of the angle ADC is 80°, and the measure of the angle BDC is 50°.\r\n" );
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document.write( "     Hence, the measure of the angle ADB is 80° - 50° = 30°.    ANSWER\r\n" );
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