Advertisements
Advertisements
प्रश्न
In the given figure, MP is the bisector of ∠P and RN is the bisector of ∠R of parallelogram PQRS. Prove that PMRN is a parallelogram.
Advertisements
उत्तर
Construction: Join PR.
Proof:
∠QPM = `(1)/(2)∠"P"` ....(AP is the bisector of ∠P)
∠SRN = `(1)/(2)∠"R"` ....(RN is the bisector of ∠R)
⇒ ∠QPM = ∠SRN (i)....[∠P =∠R (Opposite angles of a parallelogram)]
Now, ∠QPR = ∠SRN (ii)....[Alternate angles since PQ || SR)
Subtracting (ii) from (i), we get
∠QPM - ∠QPR = ∠SRN - ∠SRP
⇒ ∠RPM =∠PRN
⇒ PM || RN ...(Alternate angles are equal)
Similarly, RM || PN
Hence, PMPN is a parallelogram.
APPEARS IN
संबंधित प्रश्न
E is the mid-point of side AB and F is the mid-point of side DC of parallelogram ABCD. Prove that AEFD is a parallelogram.
The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.
In the given figure, ABCD is a parallelogram.
Prove that: AB = 2 BC.
The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. 
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90o.
Points M and N are taken on the diagonal AC of a parallelogram ABCD such that AM = CN. Prove that BMDN is a parallelogram.
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: BP bisects ∠ABC.
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that: BC = BE.
Find the perimeter of the parallelogram PQRS.
In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Show that the area of triangle QPO is `9/8` of the area of the parallelogram ABCD
