Question

The diagonals of a cyclic quadrilateral intersect at the centre of a circle containing the quadrilateral, then this quadrilateral is a

• A. parallelogram
• B. rhombus
• C. rectangle
• D. square

Answer 1

The correct option is C rectangle

In given figure ABCD is a cyclic quadrilateral.

In $△$AOB and $△$OBC;

$\Rightarrow$ AO = OC and BO = DO ( Radius of circle)

$\therefore$ $△$AOB $\cong$ $△$OBC

So, $\angle$AOB = $\angle$BOC ----- (1)

$\Rightarrow$ $\angle$AOB + $\angle$BOC = 180${}^{\circ }$ (Linear pair)

$\Rightarrow$ 2$\angle$AOB = 180${}^{\circ }$ ( From (1) )

$\Rightarrow$ $\angle$AOB = 90${}^{\circ }$

Now, in $△$AOB;

$\Rightarrow$ AO = OB (Radius of circle)

$\Rightarrow$ So, $\angle$OAB = $\angle$OBA --- (2)

$\Rightarrow$ $\angle$OAB + $\angle$OBA = 90${}^{\circ }$

$\Rightarrow$ 2$\angle$OAB = 90${}^{\circ }$

$\Rightarrow$ $\angle$OAB = 45${}^{\circ }$.

Similarly, it can be proved that $\angle$OBC = $\angle$OCB = $\angle$OCD = $\angle$ODC = 45${}^{\circ }$

This means that,

$\Rightarrow$ $\angle$OBA + $\angle$OBC = 90${}^{\circ }$

Similarly, it can be proven that all angles of quadrilateral ABCD are right angles.

Hence, quadrilateral ABCD is a rectangle.