Diagonal BD of trapezoid ABCD is perpendicular to side AB and angle BAD = 40. Assuming that the smaller

| August 11, 2021 | education

Diagonal BD of trapezoid ABCD is perpendicular to side AB and angle BAD = 40. Assuming that the smaller base of the trapezoid is equal to its second lateral side, find other corners of the trapezoid.

By condition, BD is perpendicular to AB, then triangle ABD is rectangular, and its angle ADB = 180 – 90 – 40 = 50.

The small base BC, according to the condition, is equal to the lateral side of the CD, which means that the triangle BD is isosceles and the angle CBD = CDB.

The angle ABD will be equal to the angle CBD as a cross lying angle at the intersection of parallel straight lines BC and AD secant BD, then the angle ADB = BDC = CBD = 50.

Angle ADC = ADB + BDC = 50 + 50 = 100.

Angle ABC = ABD + CBD = 90 + 50 = 140.

Angle ВСD = 180 – CBD – ВСD = 180 – 50 – 50 = 80.

Answer: The angles of the trapezoid are 40, 140, 80, 100.



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