# In an isosceles trapezoid ABCD, the diagonal AC is the bisector of angle A and forms an angle CD with the lateral side, equal to 63 °.

**In an isosceles trapezoid ABCD, the diagonal AC is the bisector of angle A and forms an angle CD with the lateral side, equal to 63 °. Find the corners of the trapezoid.**

Let the value of the angle СAD = X0, then the angle BAC = X0, since AC is the bisector of angle A.

Angle ACB = CAD = X0, as the angles lying crosswise at the intersection of parallel straight lines BC and AD secant AC. Angle BAD = BAC + CAD = 2 * X0.

In an isosceles trapezoid, the angles at the base are equal, then the angle CDA = BAD = 2 * X0.

The sum of the angles at the side of the trapezoid is 180, then:

Angle ADC + ACD + ACB = 180.

2 * X + 63 + X = 180.

3 * X = 180 – 63 = 117.

X = 117/3 = 39.

Angle BAD = ADC = 2 * 39 = 78.

Angle ABC = BCD = 39 + 63 = 102.

Answer: The angles of the trapezoid are 78 and 112.