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प्रश्न
Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to each of the parallel sides, and is equal to half the difference of these sides.
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उत्तर
Join AC and BD. M and N are mid-points of AC and BD respectively. Join MN. Draw a line CN cutting AB at E.
Now, in Δs DBC and BNE,
DN = NB ...(N is the mid-point of BD, given)
∠CDB = ∠EBN ...(Alternate angles as DC || AB)
∠DNC = BNE ...(Vertically opposite angles)
⇒ ΔDNC ≅ ΔBNE ...(By A-S-A Test)
⇒ DC = BE
By Mid-point Theorem, in ΔACE, M and N are mid-points
MN = `(1)/(2)"AE" and "MN" || "AE" or "MN" || "AB"`
Also, AB || CD, therefore, MN || CD
⇒ MN = `(1)/(2)["AB" = "BE"]`
⇒ MN = `(1)/(2)["AB" = "CD"]` ...(since BE = CD)
⇒ MN = `(1)/(2)` x Difference of parallel sides AB and CD.
संबंधित प्रश्न
ABCD is a parallelogram. P and T are points on AB and DC respectively and AP = CT. Prove that PT and BD bisect each other.
ABCD is a rectangle with ∠ADB = 55°, calculate ∠ABD.
Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
P is a point on side KN of a parallelogram KLMN such that KP : PN is 1 : 2. Q is a point on side LM such that LQ : MQ is 2 : 1. Prove that KQMP is a parallelogram.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
PMRN is a parallelogram.
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SA ‖ QB and SA = QB.
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SAQB is a parallelogram.
PQRS is a parallelogram and O is any point in its interior. Prove that: area(ΔPOQ) + area(ΔROS) - area(ΔQOR) + area(ΔSOP) = `(1)/(2)`area(|| gm PQRS)
In ΔPQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that: ΔPMR = `(1)/(8)Δ"PQR"`.
The medians QM and RN of ΔPQR intersect at O. Prove that: area of ΔROQ = area of quadrilateral PMON.
