Advertisements
Advertisements
Question
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SAQB is a parallelogram.
Advertisements
Solution
Construction:
Join BS and AQ.
Join diagonal QS.
Since diagonals of a parallelogram bisect each other.
∴ OP = OR and OQ = OS
Also, PA = AB = BR
Now, OP = OR and PA = PB
⇒ OP - PA = OR - PB
⇒ OA = OB
Thus, in quadrilateral SAQB, we have
OQ = OS and OA = OB
⇒ Diagonals of a quadrilateral SAQB bisect each other.
⇒ SAQB is a parallelogram.
RELATED QUESTIONS
ABCD is a parallelogram. P and Q are mid-points of AB and CD. Prove that APCQ is also a parallelogram.
PQRS is a parallelogram. PQ is produced to T so that PQ = QT. Prove that PQ = QT. Prove that ST bisects QR.
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 the given figure, PQRS is a trapezium in which PQ ‖ SR and PS = QR. Prove that: ∠PSR = ∠QRS and ∠SPQ = ∠RQP
Prove that the diagonals of a kite intersect each other at right angles.
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 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.
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.
