From $ \measuredangle (BC,DE) $ $ = $ $ \measuredangle BAC $ $ = $ $ \measuredangle (DF,BC) $ $ \Longrightarrow $ $ DE, $ $ DF $ are symmetry WRT $ BC $, so $ U $ $ \equiv $ $ DF $ $ \cap $ $ AC, $ $ V $ $ \equiv $ $ DE $ $ \cap $ $ AB $ is the reflection of $ P, $ $ Q $ in $ BC $, respectively $ \Longrightarrow $ $ BC, $ $ PQ, $ $ UV $ are concurrent, hence from Desargues' theorem (for $ \triangle BPV, $ $ \triangle CQU $) we conclude that $ BP $ $ \cap $ $ CQ $ lies on $ AD $. i.e. $ AD, $ $ BP, $ $ CQ $ are concurrent

Let the reflection of $E$ , $F$ over $(BC)$ be $E'$ , $F'$ Let $ (FD) \cap (AC) \equiv Y $ , $(ED) \cap (AB) \equiv X $ and $(CF) \cap (BE) \equiv O$ . $CDF+CDE'=180 <=> CED+CDF=180$ wich is true , so $D,E',F$ collinear , similarily $D,F',E$ collinear. So the reflection of $Q,P$ over $(BC)$ are $X,Y$ . Now by symmetry $AD , BP , CQ $ concurrent <=> $A'D, BY , CX$ concurrent . However by Pappus on $FBX,ECY$ we get that $O,D, (CX) \cap (BY) $ are collinear . So it suffice to prove that $A',D,O$ are collinear . Now its well known that $D$ is the center of spiral similarity sending $EB$ to $CF$ hence $EC$ to $BF$ , and so $CEOD$ is cyclic . It follows that $CDA'+CDO=ADC+CDO=ADC+180-CEB=180$ . QED.

Let $FD$ intersect $A'C$ at $T$ and $ED$ intersect $A'B$ at $R$. Because $ABDE$ and $ACDF$ are cyclic, we have that $\angle CDE = \angle CAB = \angle BDF = \angle CDT$. Also, $\angle DCT = \angle DCE$, which means that $E$ and $T$ are symmetric wrt $BC$. We get that $F$ and $R$ are symmetric wrt $BC$ in a similar way. Now let us denote by $S$ the point of intersection of lines $BT$ and $CR$. Also, let $BT$ meet $\odot ABE$ at a point $U$ other than $B$ and $RC$ meet $\odot ACD$ at $V$. Now let us prove that $S$ lies on $AD$. We have that $\angle CBT = \angle DA'T = \angle CAD$ because of symmetry. Furthermore $\angle CAD = \angle CVD$. Also, $\angle BAD = \angle DCF = \angle DCR = \angle DAV$. That means that $\angle BDU = \angle CDV$, so $U, D$ and $V$ are collinear and $BUCV$ is cyclic. Now $SB \cdot SU = SC \cdot SV$ so $S$ lies on the radical axis of $\odot ABE$ and $\odot ACD$. Finally, by Pappus theorem $S$, $D$ and the intersection point of $BP$ and $CQ$ are collinear, which is what we needed to prove.

Similar to #2. Applying Desargues' Theorem for $\triangle PDA$ and $\triangle BAC$, note that line $UV$ is the reflection of $PQ$ over $BC$, so $PQ\cap BC\in UV, AC\cap DQ\equiv U$ and $AB\cap DP=V$ are collinear. Thus establishing the statement, as desired.

My solutin(basically the same as the other solutions): Let $P'\equiv DQ\cap AC,Q'\equiv DP\cap AB$.$DE,DF$ are symmetric wrt to $BC$(since $\angle{FDB}\equiv\angle{BAC}\equiv\angle{EDC}$),so by symmetry $P',Q'$ are the symmetric points of $P,Q$ wrt $BC$.Hence $PQ$ and $P'Q'$ are concurrent at a point on $BC$.Now the problem is trivial by Desargues' Theorem in $\triangle{DPQ}$ and $\triangle{ABC}$.

Here is general problem. Let $ABC$ be a triangle and $P$ is a point inside triangle. Circle $(PAB),(PAC)$ cut $CA,AB$ again at $E,F$. $Q$ is reflection of $A$ through external bisector of $\angle BPC$. $QC,QB$ cuts $PE,PF$ at $M,N$, reps. Prove that $BM,CN$ and $AP$ are concurrent.

an overkilling solution using isogonals: let's first introduce a very usefull lemma on isogonal lines: $LEMMA: $ if in triangle $ABC$. $(AP,AQ) ; (AR,AS)$ are pairs of isogonal lines wrt $\angle BAC$, and $X=PR\cap QS; Y=PS\cap RQ$ then $(AX,AY)$ are isogonal wrt $\angle BAC$ the proof is based on using the trig formula of harmonic division. now back to the problem, we consider the triangle $QDP$, and we define $N=QC\cap BP$. note that $\angle BDQ=\angle CDE=\angle BAC \rightarrow (DB,DC)$ are isogonals wrt $QDP$ on the other hand clearly $(DQ,DP)$ are isogonal wrt $\angle QDP$. applying our lemma yellds: $(DN,DA')$ are isogonal wrt $\angle ADP$, hence forth $\angle CDN=\angle A'DC=\angle CDA \rightarrow A,N,D$ are collinear, as desired.

Solution using trigonometry: Let $\angle PBC=x,\angle QCB=y$. Let $XC\cap ACD=W,XB\cap ABD=U$. We have that $AWXU$ is cyclic. Since $\angle AWU=\angle AXU,\angle AUW=\angle AXW$ and $\angle UXW+\angle UAW=180$ it's sufficent to prove $\frac{sin(\angle AWU)}{sin(\angle AUW)}=\frac{sin(\angle DXB)}{sin(\angle DXC)}$. It is equivalent to: $\frac{AU}{AW}=\frac{sin(\angle DXB)}{sin(\angle DXC)}$ which is from Sine theorem on $\triangle AUB,\triangle AWC$ $\frac{sin(\angle DXB)}{sin(\angle DXC)}=\frac{sin(x+b)sin(c)}{sin(y+c)sin(b)}$. equvalent to $ \frac{sin(x+b)sin(c)}{sin(y+c)sin(b)}=\frac{BDsin(x)}{CDsin(y)}$-(*). From Sine theorems on $\triangle BQC,\triangle BPC$ we have $\frac{sin(c+x)}{sin(x)}=\frac{BCsin(a-c)}{CDsin(a)}$ $\frac{sin(y+b)}{sin(y)}=\frac{BCsin(a-b)}{BDsin(a)}$ After substituting $ctg(x),ctg(y)$ (*) is equivalent to $cos(b)sin(c)-cos(c)sin(b)+sin(b)sin(c)\frac{BCsin(a-c)}{CDsin(a)}=\frac{BD}{CD}(cos(c)sin(b)-sin(c)cos(b)+\frac{BCsin(a-b)}{BDsin(a)})$ $sin(c-b)(1+\frac{BD}{CD})sin(a)=\frac{BC}{CD}(sin(b)sin(c)(sin(a-c)-sin(a-b)))$ $sin(c-b)sin(b+c)=sin(b)sin(c)(sin(2b+c)-sin(2c+b))$ which is true. Therefore $A,D,X$ are collinear.

buratinogigle wrote: Here is general problem. Let $ABC$ be a triangle and $P$ is a point inside triangle. Circle $(PAB),(PAC)$ cut $CA,AB$ again at $E,F$. $Q$ is reflection of $A$ through external bisector of $\angle BPC$. $QC,QB$ cuts $PE,PF$ at $M,N$, reps. Prove that $BM,CN$ and $AP$ are concurrent. $ Q $ can be replaced by any point on the isogonal conjugate of $ AP $ WRT $ \angle BPC $ : Let $ T $ $ \equiv $ $ BM $ $ \cap $ $ CN $. From $ \measuredangle NPC $ $ = $ $ \measuredangle BAC $ $ = $ $ \measuredangle BPM $ we get $ PM, $ $ PN $ are isogonal conjugate WRT $ \angle BPC $. From the dual of Desargues' involution theorem we get $ PB $ $ \mapsto $ $ PC, $ $ PM $ $ \mapsto $ $ PN, $ $ PQ $ $ \mapsto $ $ PT $ is an involution, so $ PQ, $ $ PT $ are isogonal conjugate WRT $ \angle BPC $ $ \Longrightarrow $ $ T $ $ \in $ $ AP $. i.e. $ AP, $ $ BM, $ $ CN $ are concurrent

Solution with only angle chase . $\angle ACB=\angle BED=\angle BQD$ so $FBQC$ is cyclic so $\angle BCQ=\angle BFQ=\angle BAD$.Now let $AD$ and $QC$ meet at $L_{1}$.So $ABL_{1}C$ is cyclic.Let $BP$ and $CQ$ meet at $L_{2}$, $\angle PBA^{'}=180-2 \cdot \angle BAC-\angle QDA^{'}$, $\angle BQC=\angle ACB+\angle DA^{'}C$ ,$\angle BAC+\angle QDA^{'} =\angle ACB+\angle DAC$ so $\angle BL_{2}Q=180-\angle A^{'}BL _{2}-\angle BQC=\angle BAC$ so $BACL _{2}$ is cyclic to and $C,L_{1},L_{2}$ are collinear so $L_{1}\equiv L_{2}$.

Employ barycentric coordinates. We have $A' = (-a^2 : 2S_C : 2S_B)$ by reflection. Let $D = (0,m,n)$ with $m+n=1$. Then \[ E = \left( ma^2 : 0 : b^2-ma^2 \right). \]To solve for coordinates of $P$, we set $P = (-a^2 : 2S_C : t)$ and thus require \[ 0 = \det \begin{pmatrix} 0 & m & n \\ ma^2 & 0 & b^2-ma^2 \\ -a^2 & a^2+b^2-c^2 & t \end{pmatrix}. \]Expanding the determinant we require \[ ma^2(n(a^2+b^2-c^2)-(b^2-ma^2)) = m \cdot ma^2 \cdot t \]thus $t = \frac 1m ( a^2-mb^2-nc^2)$, and we conclude \[ P = \left( -ma^2 : 2mS_C : a^2 - mb^2 - nc^2 \right). \]By analogy we get \[ Q = \left( -na^2 : a^2 - mb^2 - nc^2 : 2nS_B \right). \]Hence done (in under five minutes).

Select $P'\equiv DF\cap AC$ and $Q'\equiv AB\cap DE$. Note by reflection that the problem is equivalent to showing that $BP', CQ'$ and $A'D$ are concurrent lines. Invert about $A$ with power $AB\cdot AC$, composed with a reflection about the $A$-angle bisector, and denote inverses with a $^*$. We discover point $D^*$ on minor arc $BC$ in $\odot(ABC)$, and $A'^*$ as the circumcenter of $\odot(ABC)$. Put $E^*\equiv CD^*\cap AB, F^*\equiv BD^*\cap AC$, and define $P'^*\equiv \odot(AD^*F^*)\cap AB, Q'^*\equiv \odot(AD^*E^*) \cap AC$. Note that since $BP'^*\cdot BA=BD^*\cdot BF^*$, it follows that line $BC$ is the radical axis of $\odot(CD^*F^*)$ and $\odot(AP'^*C)$. Similarly, it is also the radical axis of $\odot(BD^*E^*)$ and $\odot(AQ'^*B^*)$. Additionally, $\odot(ACE^*), \odot(ABF^*)$, $\odot(BD^*E^*)$, $\odot(CD^*F^*)$, and $\odot(AA'^*D^*)$ concur by Miquel's Theorem on complete quadrilateral $\{AB,BD,DC,CA\}$. Inverting back, we discover that lines $BF$, $CE$ and $AD$ are concurrent at a point $G$. Additionally, $BFGD$ and $CDGE$ are cyclic quadrilaterals. Finally, if we let $H\equiv BP'\cap \odot(BDG)$ and $K\equiv CQ' \cap (CDG)$, then $H$ and $K$ lie on $\odot(ABC)$. Finally, we note that $BP'\cap CQ'$ has equal power with respect to $\odot(BDG)$ and $(CDG)$, hence it lies on the radical axis of the two circles, so we're done$.\:\blacksquare\:$

buratinogigle wrote: Here is general problem. Let $ABC$ be a triangle and $P$ is a point inside triangle. Circle $(PAB),(PAC)$ cut $CA,AB$ again at $E,F$. $Q$ is reflection of $A$ through external bisector of $\angle BPC$. $QC,QB$ cuts $PE,PF$ at $M,N$, reps. Prove that $BM,CN$ and $AP$ are concurrent. The nice proof of houssam9990 keep valide with slight change . we don' t need to prove the so-called lemma indeed it's isogonal property .clearly PB and PC are isogonal WRT $\widehat{MPN}$ at the same time obviously $ PM $ and $ PN $ are isogonal WRT $\widehat{MPN} $ then the previous property yields $PQ$ and $PR$ are isogonal WRT $\widehat{MPN}$ where $R$ is the intersection of $BM,CN$but $PQ$ and $PA$ are isogonal WRT $\widehat{PBC}$ then WRT $\widehat{MPN}$ therefore $PA=PR$. R.HAS

My solution was the same as the official solution #3, but here's one that only uses the fact that $BC$ bisects $\angle EDF$: Let $DE,DF$ meet $AB,AC$ at $X,Y$, and note that $X,Y$ are reflections of $P,Q$ across $BC$. Let $BY$ meet $AX$ at $T$, so it suffices to show that $BC$ bisects $\angle ADT$. Now apply dual Desargues' Involution Theorem on complete quadrilateral $ABXTCY$ with respect to $D$, implying $(DA,DT)(DB,DC)(DX,DY)$ is an involution. But as $\angle XDY$ is bisected by $BC$, this involution is the reflection across line $BC$, implying that $DT$ is the reflection of $DA$ across $BC$ as desired.

Here is my solution: Lemma Lines $A'D,BE,CF$ are concurrent. Proof Applying $\sqrt{bc}$ inversion about $A$, we see that this boils down to the fact that in a cyclic quadrilateral $ABCD$ with centre $O$ and Miquel Point $M$, points $A,D,O,M$ are con cyclic. $\blacksquare$ Let these lines concur at $X$. Then, notice that triangles $DEF,A'CB$ are perspective about $X$ and so by Desrague's theorem, lines $PQ,EF,BC$ are concurrent. Now by Desrague's theorem for $DPQ,ABC$ followed by reflection in $BC$, our conclusion holds.

lol

My solution: By simple angle chasing, we know that the reflection of $E$ in $BC$, $D$ and $F$ are collinear. Now by reflecting everything over $BC$ , we have $BC$, $PQ$, line joining $FD \cap AC$ and $ ED \cap AB$ are concurrent. Now by a simple application of Desargues' theorem, the lines $AD, BP$ and $CQ$ are concurrent (or all parallel).

Oops this is almost the same as CantonMathGuy's solution (I didn't realize this until after posting)

Menelaus works; let $T'$ be the reflection of any point $T$ in the diagram. Let $X$ be the intersection of $C'Q$ and $DA'$. Use Menelaus as following: Triangle $DA'E'$ and points $X, Q, C$ Triangle $BA'C$ and points $Q, E', D$ Combining these with $CB \cdot CD = CE' \cdot CA$, we easily get that $\frac{DX}{XA}=\frac{CD \cdot CB \cdot BD}{({CA'}^{2}-CB \cdot CD) \cdot CB}$, which is symmetric.

$\measuredangle$ denotes a directed angle. Let $X=\overline{BC}\cap\overline{PQ}$, $Y=\overline{CA}\cap\overline{FD}$, and $Z=\overline{AB}\cap\overline{DE}$. Reflect $E$ across $\overline{BC}$ to $E'$. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=purple; pen sec=magenta; pen fil=invisible; pair A, B, C, D, EE, F, Ap, Ep, P, Q, X, Y, Z; A=dir(135); B=dir(200); C=dir(340); D=foot(A, B, C)+0.1; EE=intersectionpoints(circumcircle(A, B, D), A -- C)[0]; F=intersectionpoints(circumcircle(A, C, D), A -- B)[1]; Ap=2*foot(A, B, C)-A; Ep=2*foot(EE, B, C)-EE; P=extension(C, Ap, D, EE); Q=extension(B, Ap, D, F); X=extension(D, EE, A, B); Y=extension(P, Q, B, C); Z=extension(D, F, A, C); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(Ap); dot(Ep); dot(P); dot(Q); dot(X); dot(Y); dot(Z); draw(A -- B -- C -- A, pri); draw(Y -- B, pri); draw(Z -- A, pri); draw(Z -- Ep, pri); draw(X -- EE, pri); draw(Q -- Ep, pri); draw(P -- C, pri); filldraw(circumcircle(A, B, D), fil, pri); filldraw(circumcircle(A, C, D), fil, pri); draw(A -- Ap, sec); draw(EE -- Ep, sec); draw(X -- Z, sec+dashed); draw(A -- X, pri); draw(P -- Q, pri); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", EE, N); label("$F$", F, W); label("$A'$", Ap, S); label("$E'$", Ep, S); label("$P$", P, SE); label("$Q$", Q, N); label("$Z$", X, S); label("$X$", Y, W); label("$Y$", Z, N); [/asy][/asy] We claim that $E'$ is collinear with $D$ and $F$. Since $ABDE$ and $ACDF$ are cyclic, $$\measuredangle FDC=\measuredangle FAC=\measuredangle BAC=\measuredangle BAE=\measuredangle BDE=\measuredangle CDE=\measuredangle E'DC,$$and our claim has been proven. Then, $P$ and $Y$ are reflections of each other across $\overline{BC}$, and so are $Q$ and $Z$. Since $P,X,Q$ are collinear by construction, $X,Y,Z$ are collinear, so by Desargue's Theorem, $\triangle ABC$ and $\triangle DPQ$ are perspective, as desired. $\square$

This Problem is so Beautiful. My Solution is nothing different from the others. RMM 2016 Day 1 P1 wrote: Let $ABC$ be a triangle and let $D$ be a point on the segment $BC, D\neq B$ and $D\neq C$. The circle $ABD$ meets the segment $AC$ again at an interior point $E$. The circle $ACD$ meets the segment $AB$ again at an interior point $F$. Let $A'$ be the reflection of $A$ in the line $BC$. The lines $A'C$ and $DE$ meet at $P$, and the lines $A'B$ and $DF$ meet at $Q$. Prove that the lines $AD, BP$ and $CQ$ are concurrent (or all parallel). $\angle FDB=\angle BAC=\angle EDC=\angle BDP$. Now the Problem becomes Symmetric WRT $BC$. So, Consider a Reflection $\mathcal{R}$ over $BC$. So, $\mathcal R:X=FD\cap AC\leftrightarrow P$ also $\mathcal R:Y=AB\cap DE\leftrightarrow Q$.Hence, $\mathcal R:XY\mapsto PQ$. Hence, $XY,PQ,BC$ concurs in other words $AB\cap DP,BC\cap PQ,AC\cap DQ$ are collinear. So, by Desargues Theorem $\{\triangle ABC,\triangle DPQ\}$ are perspective $\implies AD,BP,CQ$ are Concurrent. $\blacksquare$

Nice solutions! I forgot about Desargues... Here are two different approaches: Solution 1. Define $K=BE\cap CF$, $P'=DF\cap AC$, and $Q'=DE\cap AB$. Let $E'$ and $F'$ be the reflections of $E$ and $F$, respectively, on $BC$. [asy][asy] import olympiad; defaultpen(fontsize(10pt)); size(315); pair A, B, C, D, E, F, P, Q, Ap, Ep, Fp, K, Kp, Qp, Pp,T; real r=0.28; draw(unitcircle, yellow); A = dir(80); B = dir(230); C = dir(-50); D = r*B+(1-r)*C; E=intersectionpoints(circumcircle(A, B, D), A -- C)[1]; F=intersectionpoints(circumcircle(A,D,C), A -- B)[1]; K=extension(B,E,C,F); Ap=2*foot(A,B,C)-A; Ep=2*foot(E,D,C)-E; Fp=2*foot(F,B,D)-F; Kp=2*foot(K,B,C)-K; P=extension(D,E,Ap,C); Q=extension(B,Ap,D,F); Qp=2*foot(Q,B,C)-Q; Pp=2*foot(P,B,C)-P; T=extension(B,Pp,C,Qp); draw(circumcircle(A,B,D), cyan); draw(circumcircle(A,D,C), cyan); draw(circumcircle(F,B,D), magenta); draw(circumcircle(E,D,C), magenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$A'$", Ap, dir(Ap)); dot("$E'$", Ep, dir(Ep)); dot("$F'$", Fp, dir(Fp)); dot("$K$", K, dir(K)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$P'$", Pp, dir(Pp)); dot("$Q'$", Qp, dir(Qp)); draw(A--B); draw(B--C); draw(A--C); draw(B--E, dashed); draw(C--F, dashed); draw(A--D, blue); draw(Ap--B); draw(Ap--C); draw(C--Q, green); draw(B--P, red); draw(C--Qp, green); draw(B--Pp, red); draw(D--Q); draw(D--P); draw(D--Qp); draw(D--Pp); draw(Q--B); draw(Qp--B); draw(C--P); draw(C--Pp); draw(Ap--D, blue); draw(D--K,purple); [/asy][/asy] Firstly, note that $\angle BDF'=\angle FDB=\angle BAC=\angle CDP$, so $F'-D-P$ are collinear. Moreover, this yields that $P$ and $P'$ are symmetric with respect to $BC$. Similarly, the same is true for $Q$ and $Q'$. Now, as $\angle KFA=\angle CFA= \angle CDA=\angle BEC=\angle KEC$, $AFKE$ is cyclic. Thus, $\angle FKB=\angle FAC=\angle FDB$, so $FBDK$ is cyclic. Similarly, $KDCE$ is cyclic. Also, $\angle A'DC=\angle CDA=\angle BEC=\angle CDK$, so $A'-D-K$ are collinear. To conclude, note that $P'$ lies in the radical axis of $\odot FBCK$ and $\odot ABC$, as $P'D\cdot P'F=P'C\cdot P'A$, so $BP'$ would be the radical axis of these two circles. The analogous result holds for $CQ'$. Hence, for being the radical axes of $\odot FBDK$, $\odot KDCE$, and $\odot ABC$, $BP'$, $CQ'$, and $A'D$ concur. Reflecting on $BC$ yields the desired result. $\square$ Solution 2. Using the same diagram as in the previous solution, we proceed by Moving Points. Fix $\triangle ABC$ and animate $D$ along line $BC$. Let $I_{AB^-}$ be the point at infinity in the direction of the antiparallel to $AB$ with respect to $\triangle ABC$. Define $I_{AC^-}$ similarly. Note that the maps $D\mapsto P$ and $D\mapsto Q$ are projective by perspectivity from $I_{AB^-}$ and $I_{AC^-}$, respectively. Now, as $D$, $P$, and $Q$ move along lines $BC$, $AC$, and $AB$, respectively, they all have degree 1. Thus, the lines $AD$, $BP$, and $CQ$ have degree 1, also. Hence, the concurrency of these three lines has degree $3$, so it suffices to prove it for 4 special cases: $\bullet$ When $D=B$, the three lines trivially concur at $B$. $\bullet$ When $D=C$, the three lines trivially concur at $C$. $\bullet$ When $D=I_{BC}$, where $I_{BC}$ is the point at infinity on $BC$, $P$ becomes the point at infinity at the direction of $A'C$, so $BP$ is the parallel to $A'C$ through $B$. We get something analogous for $CQ$, so if the two lines meet at $T$, $\triangle TBC$ will be congruent to $\triangle ABC$, so $TA\parallel BC$. Now, it is clear that $AI_{BC}$ passes through $T$. [asy][asy] import olympiad; defaultpen(fontsize(10pt)); size(210); pair A, B, C, Ap, T; A = dir(80); B = dir(220); C = dir(-40); Ap=2*foot(A,B,C)-A; T=B+C-Ap; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$T$", T, dir(T)); dot("$A'$", Ap, dir(Ap)); draw(A--B); draw(B--C); draw(A--C); draw(Ap--B); draw(Ap--C); draw(T--C, green); draw(T--B, red); draw(T--A, blue); [/asy][/asy] $\bullet$ When $D$ is the foot from $A$ to $BC$, $K$ becomes the orthocenter, $E$ and $F$ become feet, and $A-K-D-A'$ become collinear. Note that, in this case, that $(A, B; F, Q')=(A, C; E, P')=-1$, so, by definition of harmonic bundle and Ceva's Theorem, $AD$, $BP'$, and $CQ'$ concur. Reflecting on $BC$ yields the result. Hence, $AD$, $BP$, and $CQ$ always concur, so we are done. $\square$

similar to evan Apply Barycentric coordinates on $\triangle ABC$. First, it is well known that $A'=(-a^2 : 2S_C : 2S_B)$ by reflection. Additionally, by PoP we have $E = \left( ma^2 : 0 : b^2-ma^2 \right)$. Now, since $P$ lies on $CA'$, it can be parameterized as $(-a^2 : 2S_C : t)$ for some $t$. We now can solve for $t$ using the intersection, as follows: $$\left|\begin{matrix} ma^2 & 0 & b^2-ma^2 \\ 0 & m & n \\ -a^2 & a^2+b^2-c^2 & t \end{matrix}\right|=0\rightarrow t = \frac{a^2-mb^2-nc^2}{m}.$$Thus, $$P = \left( -ma^2 : 2mS_C : a^2 - mb^2 - nc^2 \right)$$and $$Q = \left( -na^2 : a^2 - mb^2 - nc^2 : 2nS_B \right)$$by symmetry. The required lines clearly concur by Ceva's. $\blacksquare$

Let $E'=DF\cap A'C$ and $F'=DE\cap A'B$. Note that $D$ is the Miquel point mapping $CE'\to F'B$, so $X = BE'\cap CF'$ lies on $(DCE')$ and $(DBF')$. Now if $R=(DCE')\cap AC$ and $S = (DBF')\cap AB$, we have $\measuredangle ARD = \measuredangle CBA$ and $\measuredangle ASD = \measuredangle BCA$, so $D$ lies on $RS$ and $BCRS$ is cyclic. In particular, $A$ has equal power wrt $(DCE')$ and $(DBF')$ and hence lies on their radical axis $XD$. Pappus on the collinear triples $B,F',Q$ and $C,E',P$ finishes.

We use moving points, animating $D$ linearly. Since $\overline{DE}$ is parallel to a fixed line (antiparallel to $\overline{AB}$), $P$ (likewise $Q$) moves linearly as well, so the desired concurrency has degree $1+1+1=3$ and we only need to check four cases (drop the condition that $D$ lies on $\overline{BC}$ with $D \neq B,C$). Three of them are below. When $D=B$, $F=Q=B$, so the three lines are concurrent at $B$. Likewise, when $D=C$, the three lines are concurrent at $C$. When $D=\infty_{\overline{BC}}$, $E$ and $F$ are the points at infinity along their respective lines, so $\overline{BP} \cap \overline{CQ}$ is the point $T$ such that $ATBC$ is an isosceles trapezoid, which clearly lies on $\overline{AD}$. For the fourth, we will take $D$ as the foot of the altitude from $A$ to $\overline{BC}$, so $DEF$ is the orthic triangle. Note that in general, a quick angle chase shows that lines $\overline{DE}$ and $\overline{DF}$ are symmetric about $\overline{BC}$. By reflecting everything over $\overline{BC}$, it suffices to show that if $P'=\overline{AC} \cap \overline{DF}$ and $Q'=\overline{AB} \cap \overline{DE}$ then $\overline{AD},\overline{BP'},\overline{CQ'}$ are concurrent. But, using signed lengths, we have (by Ceva-Menelaus) $\frac{P'A}{P'C}=-\frac{EA}{EC}$ and $\frac{Q'A}{QC}=-\frac{FA}{FB}$, so $$\frac{DB}{DC}\cdot \frac{P'C}{P'A}\cdot \frac{Q'A}{QC}=\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1,$$since $\overline{AD},\overline{BE},\overline{CF}$ concur. $\blacksquare$

Here's an elementary solution using radical axis, inspired by spoiler the hidden cyclic hexagon configuration from IMO 2019/2, among others. Define the following points: $R=(ABD) \cap \overline{BP}$, $S=(ACD) \cap \overline{CQ}$ $X=(BDF) \cap \overline{BA'}$, $Y=(CDE) \cap \overline{CA'}$ $Z=(BDF) \cap (CDE)$. Claim: $BXYC$ is cyclic. Proof: It suffices to prove that $A'$, $D$, and $Z$ are collinear by power of a point at $A'$. We have \[\angle BZD+\angle DZE=\angle BFD+180^\circ-\angle DCA=\angle BFD+\angle AFD=180^\circ,\]so $B$, $Z$, and $E$ are collinear, and analogously, $C$, $Z$, and $F$ are collinear. Thus, \[\angle BDA'+\angle BDZ=\angle BDA+180^\circ-\angle CDZ=\angle BEA+\angle CEZ=180^\circ,\]so $A'$, $D$, and $Z$ are collinear, as desired. $\square$ Claim: $BXSRYC$ is cyclic. Proof: By power of a point at $P$, $BRYC$ is cyclic. By power of a point at $Q$, $BXSC$ is cyclic. Thus, $R$ and $S$ lie on $(BXYC)$, as desired. $\square$ We finish by radical axis on $(ABD)$, $(ACD)$, and $(BSRC)$. $\square$