Question
Draw a circle of radius $3\ cm$ and draw its diameter and label it as $\text{AC}$. Construct its perpendicular bisector and let it intersect the circle at $B$ and $D$. What type of quadrilateral is $\text{ABCD}$? Justify your answer.
✓
Answer
Steps of Construction:
$1.$ Taking centre $\text{OC} = 3\ cm$, draw a circle.
$2.$ Join $A$ to $C$ and draw a perpendicular bisector of $\text{AC}$ that cuts the circumference of circle at $B$ and $D$.
$3.$ Join $B$ and $D$ .
$4.$ Thus, $\text{ABCD}$ is a cyclic quadrilateral.
Justification:
In cyclic quadrilateral,
$\angle\text{B}=\angle\text{D}=90^\circ$
$\angle\text{A}=\angle\text{C}=90^\circ [$sum of co$-$interior angles$]$
$\angle\text{D}=\angle\text{B}=180^\circ$
and $\angle\text{A}=\angle\text{C}=180^\circ$
Since, opposite angles are supplementary, thus quadrilateral is a cyclic quadrilateral.
$1.$ Taking centre $\text{OC} = 3\ cm$, draw a circle.
$2.$ Join $A$ to $C$ and draw a perpendicular bisector of $\text{AC}$ that cuts the circumference of circle at $B$ and $D$.
$3.$ Join $B$ and $D$ .
$4.$ Thus, $\text{ABCD}$ is a cyclic quadrilateral.
Justification:
In cyclic quadrilateral,
$\angle\text{B}=\angle\text{D}=90^\circ$
$\angle\text{A}=\angle\text{C}=90^\circ [$sum of co$-$interior angles$]$
$\angle\text{D}=\angle\text{B}=180^\circ$
and $\angle\text{A}=\angle\text{C}=180^\circ$
Since, opposite angles are supplementary, thus quadrilateral is a cyclic quadrilateral.
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