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

Given ABCD is a parallelogram.
To prove that points P, Q, C and D are concyclic.

Construction: Join PQ

Proof:
$\angle 1=\angle A$ [exterior angle property, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.]
But $\angle A=\angle C$ [opposite angles of a parallelogram]
$\therefore \angle 1=\angle C$.......(i) [from both of the above statements]

But $\angle C+\angle D={180}^{\circ }$ [sum of co-interior angles on same side is ${180}^{\circ }$]
$\Rightarrow \angle 1+\angle D={180}^{\circ }$ [from Eq. (i)]

Thus, the quadrilateral QCDP is cyclic because sum of the opposite angles is ${180}^{\circ }$.
So, the points P, Q, C, and D are concyclic.