Notice that $\angle BXD = 90^{\circ}$ then $BX \parallel AC$. Analogously $CY \parallel AB$. Let $P = BX \cap CY$. Therefore, the triangles $ASR$ and $PYX$ are homothetic $\Rightarrow$ $AP$, $SY$ and $RX$ are concur at $D$. Since $ABPC$ is a parallelogram, $D$ is midpoint of $BC$ $\blacksquare$

AlanLG wrote: In an acute-angled triangle $ABC$, let $D$ be a point on the segment $BC$. Let $R$ and $S$ be the feet of the perpendiculars from $D$ to $AC$ and $AB$, respectively. The line $DR$ intersects the circumcircle of $BDS$ at $X$, with $X \neq D$. Similarly, the line $DS$ intersects the circumcircle of $CDR$ at $Y$, with $Y \neq D$. Prove that if $XY$ is parallel to $RS$, then $D$ is the midpoint of $BC$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Since $BSDX$ and $CRDY$ are cyclics: $$\angle BXD=\angle DYC=90$$$$\Rightarrow CY//AS \text{ and }BX//AR...(I)$$$$\angle BXS=\angle BDS=\angle YDC=\angle YRC$$Analogously: $$\angle BSX=\angle RYC$$$$\Rightarrow \triangle BSX \sim \triangle CRY...(II)$$By $(I):$ $$\Rightarrow SX//RY$$$$\Rightarrow SRXY\text{ is a parallelogram}$$$$\Rightarrow SX=RY$$By $(II):$ $$\Rightarrow BD=DC_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

I’m the author of this problem, Santiago Rodriguez from Colombia. I don’t remember exactly how I invented the problem. I just remember that I it discovered when I tried to incorporate the following “lemma” into a problem.

”Lemma”

Although as far as I know, the lemma turned out to be completely useless when solving this problem.

$$\triangle BSD\sim \triangle CYD \rightarrow \frac{BD}{DC}=\frac{SD}{DY}$$$$\triangle BDX\sim \triangle CDR\rightarrow\frac{BD}{DC} =\frac{DX}{DR} $$$\Rightarrow \left(\frac{BD}{DC}\right)^2=\frac{SD}{DY}\cdot\frac{DX}{DR}=1$(since $XY\parallel RS$) $\hspace{19cm}\blacksquare$

$XY//RS\Longleftrightarrow \dfrac{SD}{DY}=\dfrac{RD}{DX}$ Consider that $\dfrac{SD}{DY}=\dfrac{BD}{DC},\dfrac{RD}{DC}=\dfrac{1}{\dfrac{BD}{DC}}$ Hence $(\dfrac{BD}{DC})^{2}=1\Longleftrightarrow BD=DC$