# Point M is marked inside the parallelogram ABCD. Prove that the sum of the areas of triangles AMD

March 25, 2021 | education

| **Point M is marked inside the parallelogram ABCD. Prove that the sum of the areas of triangles AMD and BMC is equal to half the area of the parallelogram.**

Let’s finish the height МН, then the area of the triangle АМD will be equal to:

Samd = AD * MН / 2 cm2.

Also, in the BMC triangle, we will complete the height of MK, then Svms = BC * MK / 2. Since BC = AD, then Svms = AD * MK / 2 cm2.

Then: Samd + Svms = (AD * MН / 2) + (AD * KM / 2) = AD / 2 + (KM + MН) = AD * KН / 2.

Then Savsd = AD * KН.

Then Samd + Svms = Savsd / 2, as required.

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