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Bad solution to bad problem. $M$ is the midpoint of $PQ, A'$ is the antipode of $A$ in $(ADE)$. Note that $A',E,Q$ and $A',D,P$ are collinear. By obvious similarities $$ \frac{AH}{DM}=\frac{QA}{QD}, \frac{AG}{EM}=\frac{PA}{PE}, \frac{HE}{A'D}=\frac{QE}{QD}, \frac{GD}{A'E}=\frac{PD}{PE} ...*$$Thus $$\frac{AS}{SE} \cdot \frac{DR}{RA}\overset{Ratio Lemma}=\frac{AH}{HE} \cdot \frac{AA'}{A'E} \cdot \frac{DG}{GA} \cdot \frac{DA'}{A'A}=\frac{AH}{AG} \cdot \frac{GD}{HE} \cdot \frac{A'D}{A'E} \overset{*} = \frac{QA}{QD} \cdot \frac{PE}{PA} \cdot \frac{PD}{PE} \cdot \frac{QD}{QE} \cdot \frac{A'E}{A'D} \cdot \frac{A'D}{A'E}=\frac{PD}{PA} \cdot \frac{QA}{QE}\overset{\triangle PAD \sim \triangle QAE}=1$$as desired. $\square$

Here’s a less bashy solution which relies on pascal. Let $M$ be the midpoint of $PQ$ and $F$ be the antipode of $A$ wrt $(ADE)$. Then we know that $P-D-F, G-R-F, H-S-F,Q-E-F$ are collinear. We have $$\angle FPQ=\angle DPA=90^\circ-\angle PAD=90^{\circ}-\angle QAE=\angle EQA=\angle FQP.$$Hence, $FP=FQ$ which means that $FM\perp PQ$, so $M\in (ADE)$. Let $AF\cap DE=X$. By Pascal on $MFGEDA$, we have $P-R-X$ are collinear. Similarly, $Q-S-X$ are collinear. Thus, $PR\cap QS=X$ which lies on the perpendicular bisector of $PQ$, so $XP=XQ$. This means that $\angle XPQ=\angle XQP$ or $\angle RPA=\angle SQA$. Therefore, $$\triangle APD\cup \{R\}\sim\triangle AQE\cup \{S\}\implies AR/AD=AS/AE$$which gives us the desired result $DE\parallel RS$. Remark. Indeed, points $P,Q$ needn’t have to be the excenters, they just have to be on the $\angle A$ external angle bisector. Here, $D,E$ can be defined as the feet from $P,Q$ to $AB,AC$, respectively.