An arbitrary point E was chosen inside the parallelogram ABCD. Prove that the sum of the areas of the triangles BEC

An arbitrary point E was chosen inside the parallelogram ABCD. Prove that the sum of the areas of the triangles BEC and AED is equal to half the area of the parallelogram

Draw a segment perpendicular to the sides AD and BC, passing through point E.
Parallelogram Area:
SABCD = AD * GF
AED Triangle Area:
SAED = AD * EF / 2
BEC Triangle Area:
SBEC = BC * EG / 2
AD = BC (according to the parallelogram property).
SBEC + SAED = BC * EG / 2 + AD * EF / 2 = AD * EG / 2 + AD * EF / 2 = (EG + EF) * AD / 2 = GF * AD / 2 = SABCD / 2

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