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Question

The line segment joining the mid - points of any two sides of a triangle is parallel to the third side and equal to half of it.
In the given figure, the side AC of ΔABC is produced of E such that CE=12AC. If D is the mid - points of BC and ED produced meets AB at F and CP, DQ are drawn parallel to BA, then FD =

A
12FE
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B
13FE
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C
FE
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D
14FE
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Solution

The correct option is B 13FE
In ΔABC, D is the mid - point of BC and DQBA.
Q is the mid - point of AC.
[Converse of mid-point theorem]
AQ=QC
Now, FADQPC, and AQC is the transversal such that AQ =QC and FDP is the other transversal on them.
FD =DP ….(1) [By intercept theorem]
Now, EC=12AC=QC
In EQD, C is the mid-point of EQ and CPDQ
P must be the mid-point of DE.
DP = PE
Thus, FD = DP = PE [From (1) and (2)]
Hence, FD=13FE


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