Question
$ABCD$ is a parallelogram. $A$ circle through $A, B$ is so drawn that it intersects $AD$ at $P$ and $BC$ at $Q.$ Prove that $P, Q, C$ and $D$ are concyclic.
✓
Answer
$ABCD$ is a parallelogram. A circle through $A, B$ is so drawn that intersects $AD$ at $P$ and $BC$ at $Q.$ We have to prove that $P, Q, C$ and $D$ are concyclic. Join $PQ.$
Now, side $AP$ of the cyclic quadrilateral $APQB$ produced to $D.$
$\therefore\ \text{Ext. }\angle1=\text{int.opp. }\angle\text{B}$
$\because BA || CD$ and $BC$ cuts them
$\therefore\angle\text{B}+\angle\text{C}=180^\circ$ [ $\because$ sum of int. $\angle\text{s}$ on the same side of the transversal is $180^\circ ] $ or
$\angle1+\angle\text{C}=180^\circ\ [\therefore\angle1=\angle\text{B}\ (\text{proved})]$
$\therefore PDCQ$ is cyclic quadrilateral.
Hence, the point $P, Q$, and are concyclic.
Now, side $AP$ of the cyclic quadrilateral $APQB$ produced to $D.$
$\therefore\ \text{Ext. }\angle1=\text{int.opp. }\angle\text{B}$
$\because BA || CD$ and $BC$ cuts them
$\therefore\angle\text{B}+\angle\text{C}=180^\circ$ [ $\because$ sum of int. $\angle\text{s}$ on the same side of the transversal is $180^\circ ] $ or
$\angle1+\angle\text{C}=180^\circ\ [\therefore\angle1=\angle\text{B}\ (\text{proved})]$
$\therefore PDCQ$ is cyclic quadrilateral.
Hence, the point $P, Q$, and are concyclic.
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