Question

Hey friend plz plz give answer its urgent tomorrow is my exam... in parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively show that the line segment ab and AC trisect the diagonal BD.

Answer 1

HELLO DEAR,



GIVEN THAT:-

ABCD is a parallelogram

Hence AB || CD


=> AE || FC , AB = CD
[Opposite sides of parallelogram ABCD]


=> AE = FC
[ E and F are midpoints of AB and CD]


In quadrilateral AECF,

one pair of opposite sides are equal and parallel.


AECF is a parallelogram.



=> AF || EC

[Since opposite sides of a parallelogram are
parallel]


In ΔDPC,

F is the midpoint of DC and FQ || CP

Hence,

Q is the midpoint of DQ by converse of midpoint theorem.


=> DQ = PQ -------------- (1)


Similarly,

IN ΔAQB,

E is the midpoint of AB and EP || AQ


Hence,

P is the midpoint of DQ by converse of midpoint theorem.


=> BP = PQ ----------- (2)


From -----(1) and-----(2),


we get,

BP = PQ = DQ


Hence,

the line segments AF and EC trisect the diagonal BD of parallelogram ABCD.



I HOPE ITS HELP YOU DEAR,
THANKS