E is the mid-point of the median AD of a \( \Large \triangle ABC \). If BE is extended it meets the side AC at F, then CF is equal to


A) \( \frac{AC}{3} \)

B) \( \frac{2 AC}{3} \)

C) \( \frac{AC}{2} \)

D) None of these

Correct Answer:
B) \( \frac{2 AC}{3} \)

Description for Correct answer:
Draw line segment DG parallel to BF



Then in, \( \Large \triangle ADG, \)

\( \Large EF \parallel DG \)

and AE = ED

\( \Large \therefore AF = GC \) ...(i)

Similarly, in \( \Large \triangle BCF, \)

\( \Large DG \parallel BF \)

and BD = DC

\( \Large \therefore FG = GC \) ...(ii)

From Eqs. (i) and (ii),

\( \Large CF = \frac{2}{3}AC \)

Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry








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