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$$E$$, $$F$$ are the mid points of the sides $$BC$$ and $$CA$$, respectively, of $$\triangle ABC$$ right angled at $$B$$ and of area 24. Area of the triangle $$CEF$$ is
In $$\triangle CEF$$ and $$\triangle CBA$$, $$\frac{CE}{CB} = \frac{CF}{CA} = \frac{1}{2}$$ with the common angle $$C$$, so the triangles are similar with ratio $$\frac{1}{2}$$. The ratio of their areas is $$\frac{1}{4}$$, so area of $$\triangle CEF = \frac{24}{4} = 6$$.
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