Let us prove that triangles ABC and AEF are similar.
EF is parallel to BC by condition, the angle BAC and EAF are common for the triangles.
The angle BCA is equal to the angle EFA as the corresponding angles at the intersection of parallel lines BC and EF of the secant AC.
Then the triangle ABC is similar to the triangle AEF in the first sign of similarity – two angles.
Since EF is the middle line of a triangle, they are equal to half of the base BC, then the coefficient of similarity of triangles is: K = BC / EF = BC / BC / 2 = 2.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
Sabc / Saef = 2 ^ 2.
Sabc = 4 * 4 = 16 cm2.
Answer: Sabc = 16 cm2.
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