Point M was marked inside the parallelogram ABCD. Prove that the sum of the areas of triangle ABM

| April 30, 2021 | education

Point M was marked inside the parallelogram ABCD. Prove that the sum of the areas of triangle ABM and CDM is equal to the area of triangle BCD.

From point M inside the parallelogram, draw the perpendiculars MK to the AB side and MH to the CD side.

Then MK is the height of the triangle ABM, and MH is the height of the triangle CDM.

Sawm = AB * KM / 2.

Ssdm = SD * MN / 2 = AB * MН / 2, since the opposite sides of the parallelogram are equal.

The sum of the areas of triangles ABM and CDM is equal to:

Sawm + Ssdm = (AB * KM / 2) + (AB * MН / 2) = (AB / 2) * (KM + MN) = AB * KН / 2.

The height BM of the triangle BCD is equal to the segment KH since they are perpendicular to the parallel sides AB and CD. Then Svsd = CD * ВM / 2 = AB * KН / 2 = Savm + Ssdm, which was required to locate.



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