Question

Prove that the parallelogram circumscribing a circle is a rhombus.
OR
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 1

Since ABCD is a parallelogram,
AB = CD ...(1)
BC = AD ...(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these questions, we obtain
DR + CR +BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC ...(3)
Comparing equations (1),(2) and (3), we obtain
AB = BC = CD = DAHence, ABCD is a rhombus.
OR

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. let us join the vertices of the quadrilateral ABCD to the center of the circle.
Consider $\Delta OAP$ and $\Delta OAS$,
AP =AS (Trangents from the same point)
OP =OS (Radii of the same circle)
OA =OA (Common side)
$\Delta OAP\cong \Delta OAS$ (SSS congruence criterion)
THerefore, $A\leftrightarrow A,P\leftrightarrow S,O\leftrightarrow O$
And thus, $\angle POA=\angle AOS$
$\angle 1=\angle 8$
Similarly,
$\angle 2=\angle 3\angle 4=\angle 5\angle 6=\angle 7$
$\angle 1+\angle 2+\angle 3=\angle 4+\angle 5+\angle 6+\angle 7+\angle 8={360}^{\circ }(\angle 1+\angle 8)+(\angle 2+\angle 3)+(\angle 4+\angle 5)+(\angle 6+\angle 7)={360}^{\circ }2\angle 1+2\angle 2+2\angle 5+2\angle 6={360}^{\circ }2(\angle 1+\angle 2)+2(\angle 5+\angle 6)={360}^{\circ }(\angle 1+\angle 2)+(\angle 5+\angle 6)={180}^{\circ }\angle AOB+\angle COD={180}^{\circ }$
Similarly, we can prove that $\angle BOC+\angle DOA={180}^{\circ }$
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.