Advertisements
Advertisements
प्रश्न
Prove that if the diagonals of a parallelogram are equal then it is a rectangle.
उत्तर
Let ABCD be a parallelogram.
In ΔABC and ΔDCB,
AB = DC ...(Opposite sides of a parallelogram are equal)
BC = BC ...(Common)
AC = DB ...(Given)
∴ ΔABC ≅ ΔDCB ...(By SSS Congruence rule)
⇒ ∠ABC = ∠DCB
It is known that the sum of the measures of angles on the same side of transversal is 180°.
∠ABC + ∠DCB = 180° ...(AB || CD)
⇒ ∠ABC + ∠ABC = 180°
⇒ 2∠ABC = 180°
⇒ ∠ABC = 90°
Since ABCD is a parallelogram and one of its interior angles is 90°, ABCD is a rectangle.
APPEARS IN
संबंधित प्रश्न
SN and QM are perpendiculars to the diagonal PR of parallelogram PQRS.
Prove that:
(i) ΔSNR ≅ ΔQMP
(ii) SN = QM
PQRS is a parallelogram. T is the mid-point of RS and M is a point on the diagonal PR such that MR = `(1)/(4)"PR"`. TM is joined and extended to cut QR at N. Prove that QN = RN.
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.
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.
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 a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
MN bisects QS.
ABCD is a trapezium in which side AB is parallel to side DC. P is the mid-point of side AD. IF Q is a point on the Side BC such that the segment PQ is parallel to DC, prove that PQ = `(1)/(2)("AB" + "DC")`.
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SA ‖ QB and SA = QB.
In the given figure, PQ ∥ SR ∥ MN, PS ∥ QM and SM ∥ PN. Prove that: ar. (SMNT) = ar. (PQRS).
In ΔABC, the mid-points of AB, BC and AC are P, Q and R respectively. Prove that BQRP is a parallelogram and that its area is half of ΔABC.
