Question

ABCD is a parallelogram in which E and F are mid – points of DC and AB respectively. If AE and DF intersect at X and CF and BE intersect at Y,then show that XEYF is a parallelogram. BEST ANSWER WILL BE MARKED AS BRAINLIEST. WRONG ANSWER WILL BE REPORTED.

Answer 1

Step-by-step explanation:

AE=BE=

2

1

AB and CF=DF=

2

1

CD

But, AB=CD

∴

2

1

AB=

2

1

CD⇒BE=CF

Also, BE∥CF [∵AB∥CD]

∴ BEFC is a parallelogram.

⇒BC∥EF and BE=PH ...(i)

Now, BC∥EF

⇒AD∥EF [∵BC∥AD as ABCD is a ∥

gm

]

⇒AEFD is a parallelogram

⇒AE=GP ...(ii)

But, E is the mid-point of AB.

∴AE=BE

⇒GP=PH [Using (i) and (ii)]

solution

Answer 2

Answer:

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