$AD, BE, CF$ are altitudes in $ABC$. $P,Q$ are points on lines $BC$ and $AB$ so $QP =PF$ and $R$ on $AC$ so $RP =CP$. Prove $ QDRA$ is a cyclic 
Let $AD, BE, CF$  be  the   altitudes   of  triangle $ABC$  and  $P$ be  an  arbitrary point of  side $BC$. Point  $Q$ on   the  line $AB$ is   such  that $QP =PF$   and   the  point $R$  on  the  line $AC$ is  such that $RP =CP$. Then $ QDRA$ is  a cyclic quadrilateral.

Move $P$ on a line $BC$.
Let $R'$ and $Q'$ be orthogonal projections of $P$ on $AC$ and $AB$. Then points $A,Q',D,P,R'$ lie on the circle with diameter $AP$. Clearly $R'$ and $Q'$ are respectively midpoints of $CR$ and $BQ$.  

Now homothety at $F$ with ratio $1/2$ takes $Q$ to $Q'$, then spiral similarity at $D$ and rotational angle $180^{\circ}-\gamma$ takes $Q'$ to $R'$ (it is not difficult to see that for each $P$ on $BC$ triangles $Q'DR'$ are similary with angles $90^{\circ}-\alpha$, $90^{\circ}-\beta$ and $180^{\circ}-\gamma$.) and homothety at $C$ with ratio $2$ takes $R'$ to $R$. 
So the composition of these three, which is again spiral similarity with angle $180^{\circ}-\gamma$, takes $Q$ to $R$. Now, it is not difficult to identify the center of this spiral similarity, namely $D$. So $A,Q,D$ and $R$ are conclyclic.
Solution here is pretty convoluted. Any simpler idea?
 A: I use your definition of $Q'$, $R'$, and prove that $D$, $Q'$, $A$, $R'$ are concyclic as you did. Moreover, since $\angle CDA=90^\circ=\angle CFA$, points $C$, $D$, $F$, $A$ are concyclic. Hence $\angle Q'FD = \angle R'CD$.
It follows that $\triangle Q'FD \sim \triangle R'CD$. Hence 
$$\frac{FQ'}{FD} = \frac{CR'}{CD}.$$
Since $FQ=2FQ'$ and $CR=2CR'$, it follows that 
$$\frac{FQ}{FD}=\frac{CR}{CD}.$$
Since $\angle QFD = \angle RCD$, we obtain $\triangle QFD \sim \triangle RCD$. Hence $$\angle DQF = \angle DRC.$$ 
Hence $D$, $Q$, $A$, $R$ are concyclic.
A: Another idea: it is well-known that the triangles $BDF$ and $EDC$ are similar: there is a direct similarity $\sigma$ with center $D$ such that $\sigma(B)=E$ and $\sigma(F)=C$. 
Now, the rotation angle is easily computed to be $\pi-A$; thus, the oriented angles $Q’D\sigma(Q’)$ and $Q’DR’$ are equal, hence $D,R’$ and $\sigma(Q’)$ are collinear. 
$B,F,Q’$ are collinear, hence $E=\sigma(B)$, $C=\sigma(F)$ and $\sigma(Q’)$ are collinear. Therefore $\sigma(Q’)$ lies on lines $DR’$ and $CE$, hence $\sigma(Q’)=R’$. 
It follows easily from the properties of similarities that $\sigma(Q)=\sigma(R)$, thus the angle $QDR$ is equal to the similarity angle which is $\pi-A=\pi-QAR$, therefore $QARD$ is cyclic.
