In a parallelogram ABCD, the point K is the middle of the side of the CD. It is known that KA = KB. Prove that the given parallelogram is a rectangle

Consider the triangles BCK and KDA. CK = KD, because point K is the middle of CD, KA = KB (from the condition of the problem), CB = AD (by the property of the parallelogram). Accordingly, the triangles BCK and KDA are equal (according to the third sign of the equality of the triangles).
It follows from the equality of these triangles that ∠BCK = ∠KDA.
BC || AD (by definition of a parallelogram), consider the side of the CD as secant to these parallel sides. Then it turns out that the sum of the angles BCK and KDA is 180 °, because these corners are internal one-sided. It follows that each of these angles is 90 °.
Now consider the sides AB and CD, they are parallel (also by the definition of a parallelogram). Consider side BC as secant to these parallel sides.
∠CBA and ∠KCB are internal one-way. Therefore, their sum is 180 °. And since ∠KCB = 90 °, then ∠CBA is also equal to 90 °.
It can be similarly proved that ∠DAB is also equal to 90 °.
A parallelogram in which all angles are straight (i.e. 90 °) is called a rectangle (by definition).

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