1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
13FE
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
FE
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
14FE
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 13FEIn ΔABC, D is the mid - point of BC and DQ∥BA. ∴ Q is the mid - point of AC. [Converse of mid-point theorem] ⇒AQ=QC Now, FA∥DQ∥PC, 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 CP∥DQ ∴ P must be the mid-point of DE. ∴ DP = PE Thus, FD = DP = PE [From (1) and (2)] Hence, FD=13FE

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
The Mid-Point Theorem
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program