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

| June 19, 2021 | education

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

Consider the formula for the areas of triangles. Let’s draw the heights in triangles – EH1 – the height of the triangle △ AED and – EH2 – the height △ BEC.
S △ (BEC) = 1/2 * (BC) * (EH2),
S △ AED = 1/2 * (AD) * (EH1) = 1/2 * BC * (EH2).

EH1 + EH2 = H1H2 is the height of the parallelogram ABCD, since both heights are perpendicular to the parallel sides BC and AD.
S △ (BEC) + S △ (AED) = 1/2 * [BC * (EH1 + EH2)] = 1/2 * BC * H1H2.

S (ABCD) = BC * E1E2 = AD * E1E2.
Judging by the expressions, it can be seen that the sum of the areas of the triangles △ BEC and △ AED is equal to half the area S (ABCD).



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