Mathematics
Prove that the parallelogram circumscribing a circle is a rhombus.
Circles
1 Like
Answer
ABCD is a parallelogram. Therefore, opposite sides are parallel and equal.
∴ AB || CD and BC || AD.
∴ AB = CD and BC = AD
The lengths of tangents drawn from an external point to a circle are equal.
Therefore,
⇒ BP = BQ ……….. (1)
⇒ CR = CQ ……….. (2)
⇒ DR = DS ……….. (3)
⇒ AP = AS ……….. (4)
Adding (1) + (2) + (3) + (4), we get :
⇒ BP + CR + DR + AP = BQ + CQ + DS + AS
⇒ (BP + AP) + (CR + DR) = (BQ + CQ) + (DS + AS)
⇒ AB + CD = BC + AD
Substitute CD = AB and AD = BC since ABCD is a parallelogram, then
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
∴ AB = BC = CD = DA
This implies that all the four sides are equal.
Hence, proved that the parallelogram circumscribing a circle is a rhombus.
Answered By
3 Likes
Related Questions
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively in figure. Find the sides AB and AC.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
In the given figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.