STKP parallelogram O intersection point of diagonals prove that triangle SOT = triangle KOP.

Since STKP is a parallelogram, its opposite sides are equal, therefore, ST = PK.

Let us consider the angles OCР and SOT, which are equal to each other as criss-crossing angles at the intersection of parallel lines TK and SK secant PT. ∠OРK = ∠SOT.

Let us consider the angles KOR and TSO, which are equal to each other as criss-crossing angles at the intersection of parallel lines TK and SK of the secant SK. ∠KOP = ∠TSO.

Then, according to the second criterion of equality of triangles “If the side and the angles adjacent to it of one triangle are respectively equal to the side and the angles adjacent to it of the other triangle, then such triangles are equal.”

Δ SOT = Δ KOP, as required.