Question

In the following figure, AC and BD are diameters of the circle with centre O. The quadrilateral ABCD is a ___.

• A. Rectangle
• B. Square
• C. Rhombus
• D. Kite

Answer 1

The correct option is A

Rectangle

In $△$AOD and $△$BOC,

OA=OB (radii)

OC=OD (Radii)

$\angle$AOD = $\angle$BOC ( vertically opposite angles)

$△$AOD $\cong$ $△$BOC (SAS congruency rule)

Therefore, by CPCTC, AD = BC and $\angle$ACB = $\angle$CAD.
We know that, if a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.
So, AD||BC.

Similarly, $△$AOB $\cong$ $△$COD by SAS congruency rule.
Then by CPCTC, AB = CD and $\angle$CAB = $\angle$ACD.
So, AB||DC.

This proves that ABCD is a parallelogram.

Further, consider $△$OCB,

Let $\angle$OBC = x.

OC = OB (radii)

$⟹\angle$OCB = $\angle$OBC = x
(Angles opposite to equal sides of a triangle are equal)

$\angle$COB = ${180}^{\circ }$ - $\angle$OCB - $\angle$OBC
= ${180}^{\circ }$ - 2x

$\angle$AOB = ${180}^{\circ }$ - $\angle$COB [linear pair]
= ${180}^{\circ }$ - (${180}^{\circ }$ - 2x)
= 2x

Now, in $△$AOB,

OA = OB (radii)

$\angle$OAB = $\angle$OBA = y
(Angles opposite to equal sides of a triangle are equal)

$\angle$OAB + $\angle$OBA + $\angle$AOB= ${180}^{\circ }$

$⟹$ y + y + 2x = 180${}^{\circ }$

$⟹$ x + y = 90${}^{\circ }$

Then,
$\angle$B = $\angle$OBC + $\angle$OBA = x + y = 90${}^{\circ }$

So, ABCD is a quadrilateral, with opposite sides equal and internal angles measuring 90${}^{\circ }$.
Therefore, ABCD is a rectangle.