Question

Prove that the quadrilateral $ABCD$ shown in the figure is a cycle quadrilateral.

Answer 1

Construction: Join $P$ and $R$ and $Q$ and $S$.
Let $\angle ADP=x$ and $\angle RAD=y$
Now, consider the quadrilateral $ADPR$.
It is clear from the figure that $ADPR$ is a cyclic quadrilateral and hence opposite angle are supplementary.
Thus, $m\angle ADP+m\angle ARP={180}^{\circ }$
$\Rightarrow x+m\angle ARP={180}^{\circ }\Rightarrow m\angle ARP={180}^{\circ }-x$
Similarly, $\angle RAD$ and $\angle DPR$ are supplementary.
Hence, we have $m\angle RAD+m\angle DPR={180}^{\circ }$
$\Rightarrow y+m\angle DPR={180}^{\circ }\Rightarrow m\angle DPR={180}^{\circ }-y$
Now, $m\angle DPR+m\angle RPQ={180}^{\circ }$ .... (Linear pair of angles)
$\Rightarrow {180}^{\circ }-y+m\angle RPQ={180}^{\circ }$
$\Rightarrow RPQ=y$
Similarly, $m\angle ARP+m\angle PRS={180}^{\circ }$ ..... (Linear pair of angles)
Thus, ${180}^{\circ }-x+m\angle PRS={180}^{\circ }$
$\Rightarrow \angle PRS=x$
Since, $PQSR$ is a cyclic quadrilateral, we have
$m\angle PQS={180}^{\circ }-x;m\angle RSQ={180}^{\circ }-y$
Again, $\angle PQS$ and $\angle SQC$ are linear pair,
$\Rightarrow m\angle SQC={180}^{\circ }-({180}^{\circ }-x)=x$
And, $\angle RSQ$ and $\angle QSB$ are linear pair,
$\Rightarrow m\angle QSB={180}^{\circ }-({180}^{\circ }-y)=y$
Now, $QCBS$ is a cyclic quadrilateral and hence, $\angle SQC$ and $\angle SBC$ are supplementary.
Also, $\angle QSB$ and $\angle QCB$ are supplementary.
$\Rightarrow m\angle SBC={180}^{\circ }-x$ and $m\angle QCB={180}^{\circ }-y$
Let us now consider the quadrilateral:
Here, we have $\angle ADP=x$ and $m\angle SBC={180}^{\circ }-x$
And, $\angle RAD=y$ and $m\angle QCB={180}^{\circ }-y$
As the opposite angles in the quadrilateral $ABCD$ are supplementary, it is a cyclic quadrilateral.