Trapezoid ABCD with bases AD and BC is inscribed in a circle with diameter AD.

| February 16, 2021 | education

Trapezoid ABCD with bases AD and BC is inscribed in a circle with diameter AD. Find the corners of a trapezoid if its diagonals intersect at an angle of 40 degrees.

Since the trapezoid is inscribed in a circle, this trapezoid is isosceles.

The ВOС angle is adjacent to the СОD angle, the sum of which is 180.

Then the angle BOC = AOD = 180 – COD = 180 – 40 = 140.

The triangles AOD and BOS are isosceles, since BO = CO, AO = DO, by the property of the isosceles trapezoid diagonals. Then the angle OBC = OCB = OAD = ODA = (180 – 140) / 2 = 20.

COD angle = AOB = 40.

The angle of ABD and ACD = 90, since they are based on the diameter of the circle.

Then the angle ABC = BCD = 90 + 20 = 110.

Since the sum of the angles at the lateral sides of the trapezoid is 180, then the angle DAB = CDA = 180 – 110 = 70.

Answer: The angles of the trapezoid are 70 and 110.



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