Advertisements
Advertisements
प्रश्न
Prove that the rhombus, inscribed in a circle, is a square.
Prove that the cyclic rhombus is a square.
Advertisements
उत्तर
Let ABCD be a rhombus, inscribed in a circle
Now, ∠BAD + ∠BCD
(Opposite angles of a parallelogram are equal)
And ∠BAD + ∠BCD =180°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
∴ ∠BAD + ∠BCD = `(180^circ)/2` = 90°
The other two angles are 90°, and all the sides are equal.
∴ ABCD is a square.
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC

Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate: ∠AEF

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.

In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.

In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.

In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.

