Given a square ABCD, on the side BC point K (arbitrary), prove that triangle AKD is equal to half the area of the square.

Let the side of the square ABCD be equal to a.

Let us express the area of the square in terms of a:
S square = a * a = a².

From point K, let us drop the perpendicular KН to the AD side. This perpendicular is the height of triangle AKD. The length of this perpendicular is equal to the side of the square:
KН = a.

Let’s write the area of the triangle through its base and height:
S tr = 1/2 * AD * KН = 1/2 * a * a = 1/2 * a².

Let’s find the ratio of the area of a triangle and the area of a square:
S tr / S square = (1/2 * a²) / a² = 1/2.

Answer: the area of a triangle is half the area of a square.