If $P, Q$ and $R$ are the mid-points of the sides $BC, CA$ and $AB$ of a triangle $ABC$ and $AD$ is the perpendicular from $A$ on $BC$,prove that $P, Q, R$ and $D$ are concyclic.

(N/A) We have to prove that $R, D, P$ and $Q$ are concyclic.
Join $RD, QD, PR$ and $PQ$.
Since $RP$ joins $R$ and $P$,the mid-points of $AB$ and $BC$,by the Mid-point theorem,$RP \parallel AC$.
Similarly,$PQ \parallel AB$.
Therefore,$ARPQ$ is a parallelogram. So,$\angle RAQ = \angle RPQ$ (Opposite angles of a parallelogram) ... $(1)$
In right-angled triangle $ABD$,$DR$ is the median to the hypotenuse $AB$. Therefore,$RA = RD$,which implies $\angle 1 = \angle B$.
Similarly,in right-angled triangle $ACD$,$DQ$ is the median to the hypotenuse $AC$. Therefore,$QA = QD$,which implies $\angle 3 = \angle C$.
In $\triangle ABC$,$\angle A + \angle B + \angle C = 180^{\circ}$.
Also,in $\triangle RDQ$,$\angle RDQ = 180^{\circ} - (\angle DRQ + \angle DQR) = 180^{\circ} - (2\angle B + 2\angle C) = 180^{\circ} - 2(\angle B + \angle C) = 180^{\circ} - 2(180^{\circ} - \angle A) = 2\angle A - 180^{\circ}$.
Alternatively,using the property of cyclic quadrilaterals,since $\angle RPQ = \angle A$ and $\angle RDQ = 180^{\circ} - \angle A$ is not directly applicable,we observe that $\angle RDQ = \angle RDA + \angle ADQ = \angle RAD + \angle QAD = \angle A$. Since $\angle RPQ = \angle A$ and $\angle RDQ = \angle A$,the points $R, D, P, Q$ are concyclic as they subtend equal angles on the same side of the segment $RQ$.