Advertisements
Advertisements
प्रश्न
Prove that a cyclic parallelogram is a rectangle.
Advertisements
उत्तर
Let ABCD be a cyclic parallelogram.
∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral) ...(1)
We know that opposite angles of a parallelogram are equal.
∴ ∠A = ∠C and ∠B = ∠D
From equation (1),
∠A + ∠C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2 ∠A = 180°
⇒ ∠A = 90°
Parallelogram ABCD has one of its interior angles, which is 90°. Therefore, it is a rectangle.
APPEARS IN
संबंधित प्रश्न
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
