Advertisements
Advertisements
प्रश्न
The diagonal BD of a parallelogram ABCD bisects angles B and D. Prove that ABCD is a rhombus.
Advertisements
उत्तर
Given: ABCD is a parallelogram where the diagonal BD bisects
parallelogram ABCD at angle B and D
To Prove: ABCD is a rhombus
Proof: Let us draw a parallelogram ABCD where the diagonal BD bisects the parallelogram at an angle B and D.
Construction: Let us join AC as a diagonal of the parallelogram ABCD
Since ABCD is a parallelogram in which diagonal BD bisects ∠B and ∠D
BD bisect ∠B,
∠ABD = ∠CBD = `1/2`∠ABC
BD bisect ∠D,
∠ADB = ∠CDB = `1/2`∠ABC
∴ ∠ABC = ∠ADC ...(Opposite angles of parallelogram are equal)
∴ `1/2`∠ABC = `1/2` ∠ADC
∠ABD = ∠ADB and ∠CBD = ∠CDB
∴ AD = AB and CD = BC
As ABCD is a parallelogram, opposite sides are equal
AB = CD and AD = BC
AB = BC = CD = AD
∴ ABCD is a rhombus.
APPEARS IN
संबंधित प्रश्न
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.
In the alongside diagram, ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B.
Prove that:
- AQ = BP
- PQ = CD
- ABPQ is a parallelogram.
Prove that the bisectors of opposite angles of a parallelogram are parallel.
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.
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: QR = QT
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: RT bisects angle R
PQRS is a parallelogram. T is the mid-point of PQ and ST bisects ∠PSR.
Prove that: ∠RTS = 90°
ABCD is a parallelogram. The bisector of ∠BAD meets DC at P, and AD is half of AB.
Prove that: BP bisects ∠ABC.
In the given figure, the perimeter of parallelogram PQRS is 42 cm. Find the lengths of PQ and PS.
Which of the following statement is correct?
