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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