To solve the problem, we can assume a common scenario where the point (E) is the centroid of (\triangle ABC) (the intersection of its medians). Here's the reasoning:

Key Observations:

• The centroid divides each median into a ratio (2:1) (longer segment from vertex to centroid).
• The centroid splits (\triangle ABC) into three smaller triangles of equal area.

Proof:

Let (AD) be a median (so (D) is the midpoint of (BC)), and (E) the centroid.

• (\text{Area of }\triangle ABD = \text{Area of }\triangle ADC = \frac{1}{2}\text{Area of }\triangle ABC) (since they share the same height from (A) to (BC) and equal bases (BD=DC)).
• (E) divides (AD) into (AE:ED=2:1). Thus, (\text{Area of }\triangle BED = \frac{1}{3}\text{Area of }\triangle ABD) (same base (BD), heights ratio (1:3)).
• Similarly, (\text{Area of }\triangle CED = \frac{1}{3}\text{Area of }\triangle ADC).

Adding these:
(\text{Area of }\triangle BEC = \text{Area of }\triangle BED + \text{Area of }\triangle CED = \frac{1}{3}\left(\frac{1}{2}\text{Area ABC} + \frac{1}{2}\text{Area ABC}\right) = \frac{1}{3}\text{Area ABC}).

Answer: (\boxed{\dfrac{1}{3}})

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