Consider the following lemma: Lemma: Let unique $b, y, i, j \in \mathbb P^1$ be fixed and $a \in \mathbb P^1$ be variable. There is a unique involution on $\mathbb P^1$ swapping $i \leftrightarrow j$ and $a \leftrightarrow b$. Suppose, this involution swaps $x \leftrightarrow y$. Then $a \mapsto x$ is projective. Proof: WLOG assume $i = 0$ and $j = \infty$. Then the involution is simply $t \mapsto \tfrac{ab}{t}$, so $a \mapsto x = a\mapsto \tfrac{ab}{y} = \tfrac{b}{y} \cdot a$, which is projective. $\square$ Now for the problem, let $\ell$ denote line $AM$ and animate $A$ on $\ell$. Let $I, J$ be the circular points at infinity. Clearly $A \mapsto CA$ is projective. Now by the lemma with $b = CB, y = CD, i = CI, j = CJ, a = CA$, $CA \mapsto CD'$ is projective, so $\deg CD' = 1$. Similarly, $\deg BD'=1$. Finally, $BD'$ and $CD'$ coincide when $A=D$, so by Moving Points Lemma, $\deg D' = 1 + 1 - 1 = 1$. Now $\deg DD' = 1$ and $\deg AK = 1$, so $\deg (AK \cap DD') = 2$, so the condition that this point lies at infinity also has degree $2{}$. It suffices to check the conclusion for three cases. If $A=M$, then $D'$ is the reflection of $D$ over $BC$, so $DD' \parallel MK = AK$. If $A$ is the second intersection point of $\ell$ with $(DBC)$, then $D'$ is the point at infinity along the $A$-symmedian of $\triangle ABC{}$. But this is just $AK$ since $ABCD$ is cyclic, so $DD' \parallel AK$. if $A$ is the point at infinity along $\ell$, then $\ell$ is the $A$-angle bisector of $\triangle ABC{}$. It contains $D{}$, so it contains $D'$ as well. Thus $DD' = \ell \parallel AK$. We are done. $\blacksquare$

@above moving points . Anyways, is there any way to finish in this way can anyone $\textbf{help}$? I am not being able to proceed any further . I have provided three versions of the problem as follows. The reason I bring the Brokard part in is because the configuration seems to be little identical to Brokard. I am not being able to use the fact that $A$ lies on the median in the current problem versions in any manner . Let $A''$ denote the isogonal conjugate of $A$ wrt $\triangle DBC$. Now let $A'$ denote the reflection of $A''$ over $BC$. So we now have that $\measuredangle CBA'=\measuredangle A''BC=\measuredangle DBA$ and $\measuredangle BCA'=\measuredangle A''CB=\measuredangle DCA$ and so we have that $A'\equiv D'$. So now change our reference triangle to $\triangle DBC$ and this becomes our current problem. $\textbf{PROBLEM: }$Let $DBC$ be a triangle and $M$ be the midpoint of $BC$. Let $A$ be a point on $DM$. Now $A'$ denote the point such that $\measuredangle A'BC=\measuredangle ABD$ and $\measuredangle A'CB=\measuredangle ACD$. Also, let the tangents to $\odot(DBC)$ meet at $T$. Then prove that $DA'\parallel AT$. Also, let $X=BA\cap\odot(DBC)$, $Y=CA\cap\odot(DBC)$. Now let $P=CX\cap BY$. Now notice that by Pascal's Theorem on $BBXCCY$, we have that $\overline{T-A-P}$ are collinear. So problem can also be stated as follows. $\textbf{PROBLEM: }$Let $DBC$ be a triangle and $M$ be the midpoint of $BC$. Now $A$ be a point on $DM$. Let $X=BA\cap\odot(DBC)$ and $Y=CA\cap\odot(DBC)$. Then $P=BY\cap CX$ and $A'$ be a point such that $\measuredangle A'BC=\measuredangle ABD$ and $\measuredangle A'CB=\measuredangle ACD$. Then prove that $PA\parallel DA'$. Now furthermore let $Q=XY\cap BC$ and $O$ be the center of $\odot(DBC)$. Now by Brokard's Theorem, we have that $AP\perp OQ$. Thus to prove $DA'\parallel AP$, it is equivalent of proving $DA'\perp OQ$. $\textbf{PROBLEM: }$Let $DBC$ be a triangle and $M$ be the midpoint of $BC$. Now $A$ be a point on $DM$. Let $X=BA\cap\odot(DBC)$ and $Y=CA\cap\odot(DBC)$. Then $Q=XY\cap BC$ and $A'$ be a point such that $\measuredangle A'BC=\measuredangle ABD$ and $\measuredangle A'CB=\measuredangle ACD$. Then prove that $OQ\perp DA'$.

Let $AD$ meets $\odot (BCD)$ again at $D''$, $L=DK\cap BC$, $E$ is a reflection of $D$ over $M$. Fix $B,C,D$ (and $M,K$), and animate $A$ by $DM$. Claim. $D'$ lies on reflection $\ell$ of line $AK$ over $BC$. Proof. The statement is equivalent to the fact that composition of rotation over $B$ by $\angle ABD'$, projection from $B$ on $\ell$, rotation over $C$ by $\angle D'CA$ and projection from $C$ on $DM$ is the identical homography $\overline{DM} \mapsto \overline{DM}$,which suffices to be checked for three cases. $A=D$. Then obviously $D'=L$. $A=M$. Then $D'$ is a reflection of $D$ over $BC$. $A=E$. Then $\angle CBD'=\angle CBE+\angle EBD'=\angle D'CE+\angle ECB=\angle D'CB=\angle CDB,$ so $D'$ is a reflection of $K$ over $BC$ $\Box$ By the claim mapping $A\mapsto D'$ is a homography $DM\mapsto \ell$, so it's suffice to check that lines $AK$ and $DD'$ meet on (fixed) infinite line for three cases. $A=D$. Then $D'=L$, so $A=D,K,D'$ are collinear. $A=E$. Then $DD'$ and $AK$ are symmetric wrt $M$, and so parallel. $A=D''$. Then $AK,AM$ are isogonals wrt angle $BAC$. Also $D\in \odot (ABC),$ so $D'$ is the infinite point of $AK$ implying $AK\parallel DD'$ $\blacksquare$

Vitriol wrote: Consider the following lemma: Lemma: Let unique $b, y, i, j \in \mathbb P^1$ be fixed and $a \in \mathbb P^1$ be variable. There is a unique involution on $\mathbb P^1$ swapping $i \leftrightarrow j$ and $a \leftrightarrow b$. Suppose, this involution swaps $x \leftrightarrow y$. Then $a \mapsto x$ is projective. Proof: WLOG assume $i = 0$ and $j = \infty$. Then the involution is simply $t \mapsto \tfrac{ab}{t}$, so $a \mapsto x = a\mapsto \tfrac{ab}{y} = \tfrac{b}{y} \cdot a$, which is projective. $\square$ Now for the problem, let $\ell$ denote line $AM$ and animate $A$ on $\ell$. Let $I, J$ be the circular points at infinity. Clearly $A \mapsto CA$ is projective. Now by the lemma with $b = CB, y = CD, i = CI, j = CJ, a = CA$, $CA \mapsto CD'$ is projective, so $\deg CD' = 1$. Similarly, $\deg BD'=1$. Finally, $BD'$ and $CD'$ coincide when $A=D$, so by Moving Points Lemma, $\deg D' = 1 + 1 - 1 = 1$. Now $\deg DD' = 1$ and $\deg AK = 1$, so $\deg (AK \cap DD') = 2$, so the condition that this point lies at infinity also has degree $2{}$. It suffices to check the conclusion for three cases. If $A=M$, then $D'$ is the reflection of $D$ over $BC$, so $DD' \parallel MK = AK$. If $A$ is the second intersection point of $\ell$ with $(DBC)$, then $D'$ is the point at infinity along the $A$-symmedian of $\triangle ABC{}$. But this is just $AK$ since $ABCD$ is cyclic, so $DD' \parallel AK$. if $A$ is the point at infinity along $\ell$, then $\ell$ is the $A$-angle bisector of $\triangle ABC{}$. It contains $D{}$, so it contains $D'$ as well. Thus $DD' = \ell \parallel AK$. We are done. $\blacksquare$ Brilliant. I didn't consider animating $A$ along $AM$ and instead animated $D$ along $AM$. You need to check $5$ cases to make it work there, but here it's just brilliant.

I don't see a completely synthetic solution above, so here is mine (more like a sketch). Quite hard problem if you do it without MMP. Let $K'$ be the isogonal conjugate of $K$; it is easy to see that $DD' \parallel AK$ is equivalent to $(ADD')$ being tangent to $AK'$ at $A$. Again by angle chasing, it easily follows that if $BK' \cap AC=E, CK' \cap AB=F$, then $E, F \in (BD'C)$. Let $BD' \cap (ABC)=B', CD' \cap (ABC)=C', BD \cap (ABC)=B", CD \cap (ABC)=C"$. By Pascal's theorem, $DD', B"C', B'C"$ are concurrent at $T$. We now claim that $AT$ is tangent to $(ADD')$. To prove this, we will use the following lemma: $\textbf{Lemma:}$ Let the points $P, Q$ be isogonal conjugates with respect to a triangle $ABC$. Let $BP \cap (ABC)=R, CQ \cap (ABC)=S$ and let $RS \cap PQ=T$. Then $AT$ is tangent to $(APQ)$. $\textbf{Proof:}$ Let $PQ \cap AB, AC=E, F$ and $SR \cap AB, AC=L, K$. Obviously $(APQ)$ touches $(AEF)$, so by PoP we need $TE.TF=TP.TQ$. It easy to see by applying twice Pascal that $PL \parallel AC, QK \parallel AB$, so we are easily done by Thales in $\triangle TKF, \triangle TLE$. Now, we are left to see why $A,T,K'$ are collinear. By angle chasing, $B"C' \parallel BE, B'C" \parallel CF$. Let $B"C' \cap AB=I, B'C"\cap AC=J$. By angle chasing again, $C'ID'B, B'JD'C$ are cyclic, so by Reim's theorem, $D'I \parallel AC, D'J \parallel AB$, so $AID'J$ is parallelogram and thus $AD'$ is median in $\triangle AIJ$. But $AD'$ is a symmedian in $\triangle ABC$, we get that $BIJC$ is cyclic. By Reim's again, $EF \parallel IJ$, so there is a homothety at $A$ taking $\triangle EFK'$ to $\triangle IJT$ and we are finally done!

A different synthetic approach. Solution Let $\Gamma = (BDC)$ and let $AB$ and $AC$ meet $\Gamma$ again at $B'$ and $C'$. Also let $M'$ be the midpoint of $B'C'$ and let the tangents to $\Gamma$ at $B'$ and $C'$ meet at $K'$. Notice that $A$, $K$, and $K'$ are collinear. (All three lie on the polar of $BC \cap B'C'$ with respect to $\Gamma$.) So what we want becomes $DD' \parallel KK'$. We will show that both of $DD'$ and $KK'$ are antiparallel to $MM'$ within $\angle BAC$. We begin with $DD'$. Let $AD$ meet $\Gamma$ again at $E$. Then $ABD' \sim AEC$, and so $AD' \cdot AE = AB \cdot AC$. Hence, $AD : AD' = (AD \cdot AE) : (AD' \cdot AE) = (AB \cdot AB') : (AB \cdot AC) = AB' : AC$. On the other hand, $ABCM \sim AC'B'M'$, and so $AM : AM' = AC : AB' = AD' : AD$. Furthermore, $A$, $D$, and $M$ are collinear, and the same similarity yields also $\angle BAM = \angle C'AM'$, so that $A$, $D'$, and $M'$ are collinear as well. Thus $D$, $D'$, $M$, and $M'$ are concyclic. From here, a straightforward angle chase shows that $DD'$ and $MM'$ are indeed antiparallel within $\angle BAC$, as claimed. We continue with $KK'$. Let $O$ and $R$ be the center and radius of $\Gamma$. Then $O$, $M$, and $K$ are collinear, $O$, $M'$, and $K'$ are collinear as well, and $OM \cdot OK = R^2 = OM' \cdot OK'$. Consequently, $K$, $K'$, $M$, and $M'$ are concyclic. From here, a straightforward angle chase shows that $KK'$ and $MM'$ are indeed antiparallel within $\angle BAC$, as claimed. The solution is complete. Motivation Consider two different positions of $D$ which lie on the same circle through $B$ and $C$. Since they yield the same $K$, they must also yield parallel $DD'$. But it is not obvious at all that they do. So let us figure this out first. Let $E'$ be the isogonal conjugate of $E$ with respect to triangle $ABC$. We want to show that $DD' \parallel EE'$. Some pairs of similar triangles yield $AD \cdot AE' = AB \cdot AC = AD' \cdot AE$, and the result follows. Notice now that, as $D$ varies, the slope of lines $DD'$ and $EE'$ depends only on the ratio $AD : AD' = AE : AE'$. So next up we try to rework this ratio somehow, likely with the help of the above identity between products. We rewrite it as $(AD \cdot AE) : (AD' \cdot AE)$ so that our product identity can come into play, and we observe that $AD \cdot AE$ is in fact simply the power of $A$ with respect to $\Gamma$. With this, it becomes natural to introduce $B'$ and $C'$ because they allow us to rewrite that power in different ways. We obtain a surprisingly simple expression for our ratio in terms of $A$, $B$, $C$, $B'$, and $C'$, namely $AB' : AC = AC' : AB$. Trying to make use of it, we discover that $AM : AM'$ is its reciprocal. This leads us to the solution.

Quite an amusing problem. kamatadu wrote: @above moving points . Anyways, is there any way to finish in this way can anyone $\textbf{help}$? I am not being able to proceed any further . I have provided three versions of the problem as follows. The reason I bring the Brokard part in is because the configuration seems to be little identical to Brokard. I am not being able to use the fact that $A$ lies on the median in the current problem versions in any manner . $\textbf{PROBLEM: }$Let $DBC$ be a triangle and $M$ be the midpoint of $BC$. Now $A$ be a point on $DM$. Let $X=BA\cap\odot(DBC)$ and $Y=CA\cap\odot(DBC)$. Then $Q=XY\cap BC$ and $A'$ be a point such that $\measuredangle A'BC=\measuredangle ABD$ and $\measuredangle A'CB=\measuredangle ACD$. Then prove that $OQ\perp DA'$. I don't know if you've found it or not, but $\triangle YDX \backsim \triangle BA'C$, so calling out the midpoint of $XY$ should just seal the deal for a pretty synthetic solution.

Hehehehe... If this solution actually works (please tell me it does lol), you literally just need two cases for $A$ not even 3! [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions (otherwise throws error due to the missing xmin) are done using bubu-asy.py. This adds back the dps, real xmin coordinates, changes the linewidth, the fontsize and adds the directions to the labellings. */ pair D = (-101.53003,23.24712); pair B = (-96.04827,-58.06426); pair C = (32.88824,-58.49839); pair M = (-31.58001,-58.28132); pair A = (-68.93332,-14.74513); pair Ast = (-74.91859,-42.87666); pair K = (-31.88815,-149.79821); pair Dp = (-75.02134,-73.39379); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(arc(B,14.69794,-36.09365,-0.19291)--B--cycle, linewidth(0.75) + blue); draw(arc(B,11.75835,-0.19291,35.70782)--B--cycle, linewidth(0.75) + blue); draw(arc(B,11.75835,57.95611,93.85686)--B--cycle, linewidth(0.75) + blue); draw(D--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--D, linewidth(0.5)); draw(circle((-31.42710,-12.86723), 78.85852), linewidth(0.5)); draw(B--K, linewidth(0.5)); draw(K--C, linewidth(0.5)); draw(K--A, linewidth(0.5)); draw(D--Dp, linewidth(0.5)); draw(D--K, linewidth(0.5)); draw(D--M, linewidth(0.5)); draw(B--Ast, linewidth(0.5)); draw(B--A, linewidth(0.5)); draw(Dp--B, linewidth(0.5)); dot("$D$", D, NW); dot("$B$", B, W); dot("$C$", C, dir(0)); dot("$M$", M, dir(270)); dot("$A$", A, N); dot("$A^*$", Ast, dir(0)); dot("$D'$", Dp, dir(0)); dot("$K$", K, dir(270)); [/asy][/asy] Let $A^*$ denote the isogonal conjugate of $A$ wrt $\triangle DBC$. Now let $A'$ denote the reflection of $A^*$ over $BC$. So we now have that $\measuredangle CBA'=\measuredangle A^*BC=\measuredangle DBA$ and $\measuredangle BCA'=\measuredangle A^*CB=\measuredangle DCA$ and so we have that $A'$ is the isogonal conjugate of $D$ w.r.t. $\triangle ABC$, that is $A'\equiv D'$. So now change our reference triangle to $\triangle DBC$ and the following becomes our current problem. $\textbf{PROBLEM: }$ Let $DBC$ be a triangle and $M$ be the midpoint of $BC$. Let $A$ be a point on $DM$. Now $D'$ reflection of the isogonal conjugate of $A$ w.r.t. $\triangle DBC$, over $BC$. Also, let the tangents to $\odot(DBC)$ meet at $T$. Then prove that $DD'\parallel AK$ Now for this problem, fix $D$, $B$ and $C$ and animate $A$ projectively on $DM$. Clearly the map $(1) A\mapsto BA$ is projective. Now the map $(2)$ of reflection across the angle-bisector of $\angle DBC$ is projective, and finally, the map $(3)$ of reflection over $BC$ is also projective. So composing the maps, we get that,\[A\xmapsto{(1)}BA\xmapsto{(2)}BA^*\xmapsto{(3)}BD',\]is projective. Similarly, we get that $A\mapsto CD'$ is projective. Now note that when $A\equiv D$, we get that $BD'\equiv CD'\equiv BC$. So by the Moving Points Lemma, the intersection of the moving lines $BD'\cap CD'$ has $\deg=1+1-1=1$. So it suffices to check for $1+1=2$ cases for $D$ that the intersection of $DD'\cap AK$ is a point at infinity. The following cases suffice. $A=M$. Then $D'$ is simply $=D_{BC}$ where $D_{BC}$ is the reflection of $D$ over $BC$. Clearly $DD'$ and $MK$ are both perpendicular to $BC$ thus $DD'\cap MK=\infty_{\perp BC}$. $A=\infty_{DM}$. Then let $T=DK\cap\odot(DBC)$. Now let $T'$ be the reflection of $T$ over $BC$. It is well known that $T'$ is the $D$-humpty point of $\triangle DBC$. So $T'$ lies on $AM$. Now, $\measuredangle \infty_{DM}CD=\measuredangle MDC=\measuredangle BDT=\measuredangle BCT=\measuredangle T'CB$ and similarly $\measuredangle \infty_{DM}BD=\measuredangle T'BC$. This means that $T\equiv D'$. Now clearly, $\infty_{DM}$ lies on $DM$ and also on $\infty_{DM}K$, so we are done. This solution is incomplete without the ending stretcher emoji, so here it goes...

n_bug wrote: A different synthetic approach. Solution Let $\Gamma = (BDC)$ and let $AB$ and $AC$ meet $\Gamma$ again at $B'$ and $C'$. Also let $M'$ be the midpoint of $B'C'$ and let the tangents to $\Gamma$ at $B'$ and $C'$ meet at $K'$. Notice that $A$, $K$, and $K'$ are collinear. (All three lie on the polar of $BC \cap B'C'$ with respect to $\Gamma$.) So what we want becomes $DD' \parallel KK'$. We will show that both of $DD'$ and $KK'$ are antiparallel to $MM'$ within $\angle BAC$. We begin with $DD'$. Let $AD$ meet $\Gamma$ again at $E$. Then $ABD' \sim AEC$, and so $AD' \cdot AE = AB \cdot AC$. Hence, $AD : AD' = (AD \cdot AE) : (AD' \cdot AE) = (AB \cdot AB') : (AB \cdot AC) = AB' : AC$. On the other hand, $ABCM \sim AC'B'M'$, and so $AM : AM' = AC : AB' = AD' : AD$. Furthermore, $A$, $D$, and $M$ are collinear, and the same similarity yields also $\angle BAM = \angle C'AM'$, so that $A$, $D'$, and $M'$ are collinear as well. Thus $D$, $D'$, $M$, and $M'$ are concyclic. From here, a straightforward angle chase shows that $DD'$ and $MM'$ are indeed antiparallel within $\angle BAC$, as claimed. We continue with $KK'$. Let $O$ and $R$ be the center and radius of $\Gamma$. Then $O$, $M$, and $K$ are collinear, $O$, $M'$, and $K'$ are collinear as well, and $OM \cdot OK = R^2 = OM' \cdot OK'$. Consequently, $K$, $K'$, $M$, and $M'$ are concyclic. From here, a straightforward angle chase shows that $KK'$ and $MM'$ are indeed antiparallel within $\angle BAC$, as claimed. The solution is complete. Motivation Consider two different positions of $D$ which lie on the same circle through $B$ and $C$. Since they yield the same $K$, they must also yield parallel $DD'$. But it is not obvious at all that they do. So let us figure this out first. Let $E'$ be the isogonal conjugate of $E$ with respect to triangle $ABC$. We want to show that $DD' \parallel EE'$. Some pairs of similar triangles yield $AD \cdot AE' = AB \cdot AC = AD' \cdot AE$, and the result follows. Notice now that, as $D$ varies, the slope of lines $DD'$ and $EE'$ depends only on the ratio $AD : AD' = AE : AE'$. So next up we try to rework this ratio somehow, likely with the help of the above identity between products. We rewrite it as $(AD \cdot AE) : (AD' \cdot AE)$ so that our product identity can come into play, and we observe that $AD \cdot AE$ is in fact simply the power of $A$ with respect to $\Gamma$. With this, it becomes natural to introduce $B'$ and $C'$ because they allow us to rewrite that power in different ways. We obtain a surprisingly simple expression for our ratio in terms of $A$, $B$, $C$, $B'$, and $C'$, namely $AB' : AC = AC' : AB$. Trying to make use of it, we discover that $AM : AM'$ is its reciprocal. This leads us to the solution. Great solution

Consider fixed points $D$, $B$, and $C$. Now, let $E$ be the reflection of point $D$ over line $BC$. As point $A$ moves along line $DM$, by utilizing Ceva's Theorem in trigonometric form (respectively for point $D$ with $\triangle ABC$ and point $D'$ with $\triangle EBC$) and the concept of isotomic conjugates, it becomes evident that point $D'$ navigates an isotomic line with respect to $\angle BEC$ upon $EM$. Importantly, the movements of points $A$ and $D'$ maintain a constant cross-ratio ($BA$ and $BD'$ intersect two fixed lines while keeping the same angle). Owing to this invariant cross-ratio, proving the original proposition only requires verification at merely three distinct points.