In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
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ABCD is ∥gm
AB ∥ CD
AE ∥ FC
⇒AB = CD
1 / 2 AB = 1 / 2 CD
AE = EC
AECF is ∥gm
In △DQC
F is mid point of DC
FP ∥ CQ
By converse of mid point theorem P is mid point of DQ
⇒DP = PQ (1)
∴AF and EC bisect BD
In △APB
E is mid point of AB
EQ ∥ AP
By converse of MPT ( mid point theorem )
Q is mid point of PB
⇒PQ = QB (2)
By (1) and (2)
⇒PQ = QB = DP
AF and EC bisect BD .
HOPE IT HELPS YOU !!
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