In the parallelogram ABCD, the diagonal AC splits angle A into two angles alpha (a) and beta (b)

In the parallelogram ABCD, the diagonal AC splits angle A into two angles alpha (a) and beta (b) ac = d find the area of the parallelogram.

Let’s define the angles of the parallelogram.

Angle BAD = BCD = α + β.

Angle ABC = ADC = (180 – α – β).

In the triangle ABC, the angle BCA = CAD = β as lying crosswise at the intersection of parallel lines AB and CD of the secant AC.

By the theorem of sines in a triangle ABC BC / Sinα = AC / Sin (180 – (α + β)).

BC = AC * Sinα / Sin (180 – (α + β)) = d * Sinα / Sin (α + β).

Determine the area of the triangle ABC.

Sас = АС * ВС * Sinβ / 2 = (d * d * Sinα / Sin (α + β)) * Sinβ / 2 = d2 * Sinβ * Sinα / 2 * Sin (α + β).

Since triangles ABC and ACD are equal on three sides, then Savsd = 2 * Savs = d2 * Sinβ * Sinα / Sin (α + β).

Answer: The area of the parallelogram is d2 * Sinβ * Sinα / Sin (α + β) cm2.