Question

in a parallelogram ABCD E and F are the midpoints of sides AB and CD respectively. show that AF and EC trisect the diagonal BD

Answer 1

because AB and CD are the midpoints respectively of e and f .
so,therefore 1\2 AB and CF=1\2CD ....eq.1
but ABCD is a parallelogram.
so,AB=CD and AB||CD so,1\2 AB =1\2CDA
AE =FC and AE ||DC
AECF is a parallelogram.
FAllCF ,FPllCQ.....eq.2
now in triangle PCD ,CD 's mid point of F and FPllCQ
therefore,P is the midpoint of DQ
PQ=DP.....eq.3
sim.in triangle ABP E is the midpoint of AB
BP's midpoint is Q
BQ=PQ.....eq.4
from eq.3 and4
DP=PQ=QB
hence it proved that ,
AP and CE are trisecting BD

Answer 2

Answer:

In ∆CDQ

DF = CF

DF//QC

DP = PQ -----(i) ( Converse repeating

theorem )

In ∆ABP

AE = EB ( Given )

EQ // AP ( Midpoint theorem )

PQ = QP -----(ii) ( Converse Midpoint

theorem )

From (i) and (ii)

PQ = QB = DP

Line Segments AF and EC bisect the diagonal BD