In triangle ABC, point M is the midpoint of side BC. On the straight line AM, such a point K is taken that MK

| March 8, 2021 | education

In triangle ABC, point M is the midpoint of side BC. On the straight line AM, such a point K is taken that MK = AM. Prove that a quadrilateral ABAКС is a parallelogram.

Method I.

Consider triangles BMC and СMA: BM = CM (point M – middle BC); AM = KM (by condition), the angle of the BMC = the angle of the СMA (vertical angles). So the triangles are equal.

Consequently, the ВKM angle is equal to the MAC angle. And this means that the ВC is parallel to the AC (the angles are equal as the inner ones lying crosswise).

Similarly, we prove that triangles ABM and KСA are equal and AB is parallel to KС.

A quadrilateral whose opposite sides are parallel is a parallelogram.

Method II.

Consider a quadrilateral ABKС: AK and BC are the diagonals of the quadrilateral. Point M is the middle of the BC and the middle of the AK.

By the property of a parallelogram: a quadrilateral in which the diagonals intersect and the intersection point is halved is a parallelogram.



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