Question 1210498
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Quadrilateral ABCD is a parallelogram. Let E be a point on AB, and let  F be the intersection of lines 
DE and BC. The area of triangle EBC is 4, and the area of triangle ABC is 4. 
Find the area of parallelogram ABCD.
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The post is SELF-CONTRADICTORY and describes a situation which NEVER may happen.



Indeed, triangle ABC has the base AB and the height 'h', which is the height of the parallelogram
drawn to base AB.


Triangle EBC has the base EB and the same height 'h', but EB is part of BC, shorter than BC itself.


But under this condition, the areas of triangles ABC and EBC can not be equal.


This is a contradiction, which kills the problem to the death and ruins it into dust.