Advertisements
Advertisements
प्रश्न
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Advertisements
उत्तर
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.
∴ ∠COD = 90°
Also, in rhombus, the diagonals intersect each other at 90°.
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Clearly, point O has to lie on the circle.
संबंधित प्रश्न
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
