In fact, this is true for arbitrary $D,E,F$ on the circumcircle. Sketch: Let the orthocenter of $ABC$ be $H$, of $DBC$ be $H_A$, etc. It is easy to prove with vectors that $AHH_AD$ forms a parallelogram, so the midpoints of $AH_A$ and $DH$ are the same point. But the Simson line of $A$ wrt $DBC$ passes through the midpoint of $AH_A$, and similarly for $DH$, so $A_1$ is this common intersection. Similarly we get that $A_1B_1C_1$ is a triangle homethetic to $DEF$ with homothety center $H$ and scale factor $2$.

It is a very nice problem Consider the line $m_a$.It passes through the perpendicular from $D$ to $AB,AC$.So it passes through the midpoint $A_m$ of $BC$.Similarly we may note that $l_a$ passes through the feet of $A$-altitude (say $X$) of $\triangle{ABC}$,since it is the simson line of $A$ w.r.t $\triangle{DBC}$.Now let $J$ be the interection point of $l_a$ and $DA_m$.Again by easy angle chasing(using concyclicity of quadrilaterals and all) we get $XA_1=A_1J$.Thus $XA_1=A_1J=A_1M$ and $A_1$ is the midpoint of $XJ$.Now $m_a$ is the simson line of $D$ wrt $\triangle{ABC}$ and so $m_a$ bisects $DH$ where $H$ is the orthocenter of $\triangle{ABC}$.Since $A_1$ is already the midpoint of $XJ$,we have $D,A_1,H$ collinear and moreover $A_1$ is the midpoint of $DH$.Analogously $B_1$ is the midpoint of $EH$ and $C_1$ is the midpoint of $FH$.Thus $\triangle{A_1B_1C_1} \sim \triangle{DEF}$ with factor $\frac{1}{2}$.

This problem was actually pretty "clumsy" in it's original formulation. The solution goes like this: Notice that $l_a,m_a$ are the Simson's Lines of points $A,D$ w.r.t $\triangle DBC,\triangle ABC$ and so they pass through $L_A,M_A$ where $L_A,M_A$ are the feet of $A$- altitude and the mid-point of $BC$ respectively. Now, some angle-chasing gives that $\angle A_1L_AM_A=\angle A_1M_AL_A=\frac{\angle B-\angle C}{2}$. Now, consider the homothety $\mathcal{H}$ centered at $H$ - the orthocenter of$\triangle ABC$ with ratio $+2$. Note that it sends $L_A,M_A$ to the reflection of $H$ in $BC$ and the antipode of $A$ w.r.t $\omega$ -the circumcircle of $\triangle ABC$ respectively. Now, since $H$ and $O$ are isogonal conjugates and $D$ is the mid-point of arc $BC$, we have $\mathcal{H}(H,+2): A_1 \longrightarrow D$ Proving that $\triangle A_1B_1C_1$ is similar to $\triangle DEF$ both being homothetic about $H$. The generalization suggested by Pi37 requires more insight. Anyway, it was a very nice synthetic geometry problem and was a lot fun to do.

Oops I am not familiar with simpson lines so here is my naive solution It can probably be simplified too, but by now I'm too lazy to reread it.

Nice problem. Let $H$ and $H'$ be the orthocenters of $ABC$ and $DBC$ respectively. It is well known that $AH=DH'=4R^2-BC^2$. Since $AH\parallel DH'$, $AHH'D$ is a parallelogram. So if $K$ is the midpoint of $AH'$, then $K$ is also the midpoint of $HD$. It is well known that the simon lines of $A$ and $D$ w.r.t. $DBC$ and $ABC$ pass through $K$. So $A_1=K$. Therefore, there is homothety centered at $H$ with ratio $1/2$ that sends $DEF$ to $A_1B_1C_1$. $\Box$

We consider the case where $D,E,F$ are arbitrary points on the circumcircle. The proof is essentially the same as pi37, K6160.

The Simson lines of $A,D$ wrt $\triangle DBC,\triangle ABC$ must pass through the (common) midpoint of $\overline{AH_a},\overline{DH}$ respectively (well-known, see here), so this point is $A_1$. Thus with analogous results for other vertices, a homothety centre $H$ with scale factor $2$ sends $\triangle A_1B_1C_1\mapsto\triangle DEF$.

I fail to see obvious synthetic things, so here is a mix of synthetic and complex

Denote by $M$ the midpoint of $BC$ , and by $N$ the foot of the altitude from $A$ to $BC$. With the Simson line in mind, we can prove $A_1$ lies on the nine-point circle $N_9$ by angle chasing. Also $\triangle A_1MN$ is isosceles, which means line $A_1N_9$ is perpendicular to $BC$, Thus $\triangle{A_1B_1C_1} \sim \triangle{DEF}$.

[asy][asy] unitsize(2.5inches); pair A=dir(115); pair B=dir(220); pair C=dir(-40); pair H=orthocenter(A,B,C); pair D=dir(-95); pair X=foot(A,B,C); pair XX=2X-H; pair M=foot(D,B,C); pair A1=(H+D)/2; draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(A--XX); draw(X--A1--M); draw(M--D); draw(B--D--C); dot("$A$",A,dir(135)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(180)); dot("$D$",D,dir(D)); dot("$A'$",-A,dir(-A)); dot("$X$",X,dir(45)); dot("$X'$",XX,dir(XX)); dot("$M$",M,dir(90)); dot("$A_1$",A1,dir(A1)); [/asy][/asy] Note that $\ell_A$ is the Simson line of $A$ with respect to $\triangle DBC$ and $m_A$ is the Simson line of $D$ with respect to $BC$. Thus, $\ell_A$ passes through $X$, the foot from $A$ to $BC$, and $m_A$ passes through $M$, the foot from $D$ to $BC$. Now, $XA_1$ and $MA_1$ are the two Simson lines. We claim that $A_1$ is the midpoint of $HD$. To see this, note that the midpoint of $HD$ lies on the Simson line of $D$ with respect to $ABC$, or that the midpoint of $HD$ lies on $MA_1$. We'll now show that it lies also on $XA_1$. To do this, it suffices to show that $XA_1/\parallel X'D$ where $X'$ is the reflection of $H$ in $BC$, since then the image of $XA_1$ under a homothety of scale factor $1/2$ would have to be $X'D$ (since $X\mapsto X'$). Note that $\angle A_1XM$ is the angle between the Simson lines of $D$ and $A'$ respectively, where $A'$ is the antipode of $A$. Therefore, $\angle A_1XM=\frac{1}{2}\widehat{DA'}$. We also see that the angle between $X'D$ and $BC$ is \[\frac{1}{2}(\widehat{DC}-\widehat{BX'})=\frac{1}{2}(\widehat{DC}-\widehat{CA'})=\frac{1}{2}\widehat{DA'}=\angle A_1XM,\]which proves the lines are parallel. Thus, $A_1$ is the midpoint of $HD$, so $DEF$ and $A_1B_1C_1$ are homothetic by a homothety at $H$. Note that this solution works for arbitrary $DEF$ on the circle.

Storage. Anyways Nice Lemma. USATST 2013 P1 wrote: Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $\ell_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangle $DEF$ and $A_1B_1C_1$ are similar to each other. Let $\{H,H_A\}$ be the orthocenters of $\{\triangle ABC,\triangle BDC\}$ respectively. So, $l_a$ bisects $AH_A$ bisects at a point $A_1'$ (say). and $m_a$ bisects $DH$ at point $A_1''$(say). but it's well known that $AHDH_A$ is a parallelogram. So, $AH_A,DH$ bisects each other. So, if $A_1'\equiv A_1''\equiv A_1$. So, $DA_1=A_1H$. Similarly we get that $EB_1=B_1H$ and $FC_1=C_1H$. So, $\triangle DEF$, $\triangle A_1B_1C_1$ are homothetic centered at $H$ with a scale facor of $2$. $\blacksquare$

We will prove the more general result. If $D,E,F$ are any point on $(ABC)$ instead of arc-midpoint then the desired result also hold. Assume $D,E,F$ be any points on the unit circle $(ABC)$. Let $H,J$ is the orthocenter of $\triangle ABC$ and $\triangle DBC$ respectively. We claim that $A_1$ is the midpoint of $DH$. Clearly $l_a$ and $l_b$ is the simson line of $D$ and $A$ wrt $\triangle ABC$ and $\triangle DBC$ respectively. It is well known that $l_a$ bisects $HD$. Similarly $l_m$ bisects $AJ$. Therefore the the line passes through $W$ with coordinate, \begin{eqnarray*} w=\frac{a+b+c+d}{2} \end{eqnarray*}Similarly $l_m$ also passess through $W$. So we can say $W\equiv A_1$. \vspace{2mm} Similarly we can say that $B_1, C_1$ are the midpoints of $HE, HF$ respectively. Now applying homothety at the center $H$ with factor $2$ we get $A_1\mapsto D, B_1\mapsto E, C_1\mapsto F$, which means $\triangle A_1B_1C_1\mapsto \triangle DEF$. So we conclude that $\triangle A_1B_1C_1\sim \triangle DEF$.

Let $H$ and $H_1$ be the orthocenters of $\triangle ABC$ and $\triangle DBC,$ respectively. Claim: $A_1$ is the midpoint of $\overline{DH}.$ Proof. Redefine $A_1=m_a\cap\overline{DH}.$ Notice $m_a$ is the Simson line of $D$ wrt $\triangle ABC$ so by the Simson Line Bisection Lemma, $A_1$ is the midpoint of $\overline{DH}.$ We claim $A_1$ is also the midpoint of $\overline{AH_1}$ and therefore on $\ell_a.$ WLOG let $(ABC)$ be the unit circle and note $h=a+b+c$ and $h_1=b+c+d.$ Hence, $$a_1=\tfrac{1}{2}(d+h)=\tfrac{1}{2}(a+b+c+d)=\tfrac{1}{2}(a+h_1).$$$\blacksquare$ Thus, there is a homothety at $H$ with scale factor $2$ such that $A_1\mapsto D$ and similarly $B_1\mapsto E$ and $C_1\mapsto F.$ This implies $\triangle A_1B_1C_1\sim\triangle DEF.$ $\square$

By Simson line Lemma: $A_1$ is midpoint of $DH$.

Let $H$ be the orthocenter. We claim that $A_1$ is the midpoint of $HD$. Let $A_1'$ be the midpoint of $HD$. By Simson line bisection, $A_1$ lies on $m_a.$ It remains to show $A_1'$ lies on $\ell_a$. Let $P$ denote the foot from $A$ to $BD$ and $Q$ denote the foot from $A$ to $CD.$ Then, $$p=\frac{1}{2}(a+b+d-\frac{bd}{a}),q=\frac{1}{2}(a+c+d-\frac{cd}{a}).$$Note that $A_1'=\frac{1}{2}(a+b+c+d).$ Thus, it suffices to show that $$a+b+c+d,a+b+d-\frac{bd}{a},a+c+d-\frac{cd}{a}$$are collinear. To do this, shift to 0 and scale by -1 so its $$0,c+\frac{bd}{a},b+\frac{cd}{a}.$$Thus, it suffices to show that $$\frac{ac+bd}{ab+cd}\in R.$$However, this is clearly true by taking the conjugate and multiply top and bottom by $abcd$. Thus, $A_1'=A_1$, and similarly for the other two, so $A_1B_1C_1$ is just $DEF$ dilated by 1/2 with center $H$.

Realizing the geometric sol after 5 second I bashed it out