Question -
Answer -
Given that,
ABCD is a parallelogram. E and F are the mid-points of sides ABand CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,
ABCD is a parallelogram
, AB || CD
also, AE || FC
Now,
AB = CD (Opposite sides of parallelogram ABCD)
⇒½ AB =½ CD
⇒ AE =FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal toeach other)
AF || EC (Opposite sides of a parallelogram)
Now,
In ΔDQC,
F is mid point of side DC and FP || CQ (as AF || EC).
P is the mid-point of DQ (Converse of mid-point theorem)
⇒ DP =PQ — (i)
Similarly,
In ΔAPB,
E is midpoint of side AB and EQ || AP (as AF || EC).
Q is the mid-point of PB (Converse of mid-point theorem)
⇒ PQ =QB — (ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.
Hence Proved.