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 alongside figure shows a parallelogram ABCD in which AE = EF = FC.
Prove that:
- DE is parallel to FB
- DE = FB
- DEBF is a parallelogram.
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.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: ∠RTS = 90°
In a parallelogram ABCD, E is the midpoint of AB and DE bisects angle D. Prove that: BC = BE.
In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.
In the Figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to `bar("HE")` and `bar("FG")`?
In the following figure, ABCD and AEFG are two parallelograms. If ∠C = 55º, determine ∠F.
Construct a parallelogram POUR in which, PO = 5.5 cm, OU = 7.2 cm and ∠O = 70°.
