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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
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