Question

please give the answer correctly.... and will rewarded as brainliest...​

Answer 1

$\huge{\mathcal{\underline{\green{AnSWer}}}}$

Given:

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove:

the line segments AF and EC trisect the diagonal BD.

✍︎✍︎Proof :

In quadrilateral ABCD,

AB=CD ----------- (Given)

$\frac{1}{2} AB = \frac{1}{2} CD \\ AE = CF$

✍︎✍︎In quadrilateral AECF,

AE=CF ---------- (Given)

AE || CF --------- (Opposite sides of a parallelogram)

Hence, AECF is a parallelogram.

In △DCQ,

F is the midpoint of DC.------- (given )

FP || CQ ---------- (AECF is a parallelogram)

By converse of midpoint theorem,

P is the mid point of DQ.

DP= PQ....................1

Similarly,

In△ABP,

E is the midpoint of AB. (given )

EQ || AP ------- (AECF is a parallelogram)

By converse of midpoint theorem,

Q is the midpoint of PB.

OQ= QB....................2

From 1 and 2, we have

DP = PQ = QB.

✈︎✈︎Hence, the line segments AF and EC trisect the diagonal BD.