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 1

Step $1:$ Drawing the diagram:

$ABCD$ is a parallelogram.

A circle through $A,B$ is so drawn that it intersects $AD$ at $P$ and $BC$ at $Q$.

Join $PQ$ and marked $\angle QPD=\angle 1$

Step $2:$ Proving $P,Q,C$ and $D$ are concyclic:

From figure, $APQB$ is a cyclic quadrilateral.

$\therefore \angle B+\angle APQ=180°$ (Sum of opposite angles of cyclic quadrilateral is supplementary)…….$\left(i\right)$

$\angle 1+\angle APQ=180°$ (Linear pair of angles)…………….$\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\angle B=\angle 1$……………..$\left(iii\right)$

$ABCD$ is a parallelogram.

$\therefore \angle B=\angle D$ (opposite angles of parallelogram are equal)…………………...$\left(iv\right)$

And, $\angle C+\angle D=180°$ (adjacent angles of parallelogram are supplementary)…………$\left(v\right)$

From $\left(iii\right)$ and $\left(iv\right)$, we get

$\angle 1=\angle D$……………………..$\left(vi\right)$

From $\left(v\right)$ and $\left(vi\right)$, we get,

$\angle C+\angle 1=180°$

As, Sum of opposite angles of cyclic quadrilateral $PQCD$ is supplementary.

Hence, $P,Q,C$ and $D$ are concyclic.