Question

please solve fast it's urgent.​

Answer 1

Answer:

Since, the opposite sides of a parallelogram are parallel and equal.

∴ AB || DC

⇒ AE || FC …(1)

and AB = DC

⇒ 12AB = 12DC

⇒ AE = FC …(2)

From (1) and (2), we have

AE || PC and AE = PC

∴ ∆ECF is a parallelogram.

Now, in ∆DQC, we have F is the mid-point of DC and FP || CQ

[∵ AF || CE]

⇒ DP = PQ …(3)

[By converse of mid-point theorem] Similarly, in A BAP, E is the mid-point of AB and EQ || AP [∵AF || CE]

⇒ BQ = PQ …(4)

[By converse of mid-point theorem]

∴ From (3) and (4), we have

DP = PQ = BQ

So, the line segments AF and EC trisect the diagonal BD.

Answer 2

Answer:

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Short ,easy and good answer.

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