Question

Construct a cyclic quadrilateral ABCD such that AB$=5.5$cm $\angle$ABC$={50}^o$, $\angle$BAC$={60}^o$ and $\angle$ACD$={30}^o$.

Answer 1

$Step1:$ Draw a line segment $AB=5.5cm$.

$Step2:$ Through $A$ draw $AX$ such that $\angle BAX={60}^0$

$Step3:$ Through $B$ draw $BY$ such that $\angle ABY={50}^0$

$Step4:$Mark the point where $BY$ meets $AX$ and label it as $C$.

$Step5:$ Draw the perpendicular bisectors of $AB$ and $BC$ intersecting each other at $O$.

$Step6:$ With $O$ as centre and $OA(=OB=OC)$ as radius, draw the circumcircle of $△ABC$.

$Step7:$ From $A$, draw $AZ$ such that $\angle BAZ={80}^0$, which intersects the circle at $D$.

$Step8:$ Join $AD$ and $CD$.

Thus, $ABCD$ is the required cyclic quadrilateral.