In parallelogram ABCD, the length of the diagonal is BD = 2, the angle is C = 75 degrees.

In parallelogram ABCD, the length of the diagonal is BD = 2, the angle is C = 75 degrees. The circle circumscribed around triangle ABD touches line CD. Find the area of the parallelogram.

∠ BAD = ∠ BCD = 75 ° (opposite angles of the parallelogram). Extend the CD side and consider the ADK angle.
∠ ADK = ∠ BCD = 75 ° (corresponding angles for parallel AD and BC, secant – CD).
∠ ADK rests on arc AD:
∠ ADK = 1/2 AD = 75 ° → ͜ AD = 150 °.
The angle ABD rests on the same arc:
∠ ABD = 1/2 AD = 75 °.
Consider a triangle ABD, it is isosceles (∠ BAD = ∠ ABD = 75 °), which means:
AD = BD = 2;
∠ BDA = 180 ° – 2 * 75 ° = 30 °.
On the AD side, we lower the height BH = 1/2 * BD = 1 (the leg is opposite the angle of 30 °).
S ABCD = AD * BH = 2 * 1 = 2.
Answer: The area of the parallelogram is 2.