RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

Question : 12 of 120

Marks: +1, -0

E is the mid-point of the median AD of a ΔABC, If BE produced meets the side AC at F, then CF is equal to

${AC} /3$
${ 2 AC }/3$
${AC} /2$
None of these

Solution:

(b) We draw a line segment parallel to BF.

In ΔADG,
BF or EF || DG and AE = ED (since, E is midpoint of AD)
∴ AF = GC ... (i)
Similarly, in ΔBCF,
DG || BF and BD = DC
FG = GC ... (ii)
From Eqs. (i) and (ii),
CF=

2/3AC

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