I think $Q,R$ need to lie on the different side of $AD$.

Adding some Details to the above proof Henry_2001 wrote: In triangle $ABC$, $AB>AC.$ The bisector of $\angle BAC$ meets $BC$ at $D.$ $P$ is on line $DA,$ such that $A$ lies between $P$ and $D$. $PQ$ is tangent to $\odot(ABD)$ at $Q.$ $PR$ is tangent to $\odot(ACD)$ at $R.$ $CQ$ meets $BR$ at $K.$ The line parallel to $BC$ and passing through $K$ meets $QD,AD,RD$ at $E,L,F,$ respectively. Prove that $EL=KF.$ Claim$\odot{BCQR}$ is cylic. Proof Consider $\sqrt{AB\cdot AC}$ inversion followed by reflection across the angle bisector.We get the following problem:- Inverted Problem wrote: Let $ABC$ be a triangle.Let $AI\cap \odot{ABC}=M$.Let $P$ be a point on $AM$ such that $P$ lies on the other side of $A$ than $M$.Let $\gamma_1,\gamma_2$ be two circles through $A,P$ such that $\gamma_1,\gamma_2$ are tangent to $MB,MC$ at $Q,R$ respectively.Then $\odot{BCYZ}$ is cylic. Clearly $MB=MC$ be Fact-5 and by tangency $MQ=MR$.Hence $\odot{BCQR}$ is an isosceles triangle which is cylic.Hence inverting back we get the desired Claim.$\square$ By Radical Axis on $\odot{ABD},\odot{ACD},\odot{BCQR}$ we get $BQ,CR,AD$ are concurrent at some point say $T$.Now by DDIT on $\odot{BCQR}$ with $D$ we get $(DQ,DR),(DB,DC),(DK,DT)$ are pair of involution.Projecting onto $EF$ we get $(E,F),(\infty_{EF},\infty_{EF}),(K,L)$ are involution which clearly implies $EL=KF$.$\blacksquare$

Claim $QRCB$ is cyclic Proof. This is just angle chasing. Let $QB \cap RD = \{S\}$ and $QD \cap RC = \{T\} $ Using that $AD$ is the bisector of angle $BAC$ then $\angle BQD = \angle DRC$ so $QRTS$ is cyclic. Since $QRCB$ is cyclic then by Reim's Theorem $BC \parallel ST$ . By Radical Axis $AD , CR, BQ$ are concurrent in $X$ Let $AD$ intersect $ST$ in $U$. Then $$\frac{EL}{LF} = \frac{TU}{US} = \frac{CD}{DB}$$Note that $\bigtriangleup RBS \sim \bigtriangleup QCT$ having 2 pairs of angles equal so the same holds also for the third hence $\angle CQD = \angle BRD$ From Law of Sines in $\bigtriangleup QKE$ and $\bigtriangleup RKF$ we get using the above equality $$\frac{EK}{KF} = \frac{QK}{RK} \cdot \frac{\sin{\angle RFK}}{\sin{\angle QEK}}$$But $\angle RFK = 180^{\circ} - \angle RDC$ and $\angle QEK = 180^{\circ} - \angle QDB$ so $$\frac{\sin{\angle RFK}}{\sin{\angle QEK}} = \frac{\sin{\angle RDC}}{\sin{\angle QDB}} = \frac{\frac{RC \cdot \sin{\angle DRC}}{DC}}{\frac{QB \cdot \sin{\angle BQD}}{BD}} = \frac{BD}{DC} \cdot \frac{RC}{BQ} = \frac{BD}{DC} \cdot \frac{RK}{QK}$$So $\frac{EK}{KF} = \frac {BD}{DC} = \frac{LF}{EL} $ Since $AB > AC$ the order of the points on the line parallel to $BC$ is $E,L,K,F$. Then in particular $EK + KF = EF$ and $LF + EL = EF$ so $\frac{EK}{EF} = \frac{LF}{EF}$ $\Rightarrow$ $EK = LF$.

First we relabel points. Replace the labels $P, Q, R, K, E, L, F$ with $E, F, G, J, H, K, I$ respectively. So we now want $HK = JI.$ It's easy to obtain that $\frac{AF}{FD} = \frac{AG}{GD}.$ By Sine Law in $\triangle DHI$ we have that $\frac{HK}{KI} = \frac{DH}{HI} \cdot \frac{\sin \angle HDK}{\sin \angle IDK}.$ We have by Trig Ceva in $\triangle DFG$ with point $E$ that $\frac{\sin \angle HDK}{\sin \angle IDK} = \frac{\sin \angle DFE}{\sin \angle DGE}$, which implies $\frac{\sin \angle HDK}{\sin \angle IDK} = \frac{DF}{DG} \cdot \frac{AC}{AB}.$ Combining the previous two results, we have: $$\frac{HK}{KI} = \frac{DH}{DI} \cdot \frac{DF}{DG} \cdot \frac{AC}{AB}.$$ Now, we have that $HJ = CD \cdot \frac{FH}{FD}$ and $IJ = BD \cdot \frac{GI}{GD},$ and so: $$\frac{HJ}{JI} = \frac{CD}{AB} \cdot \frac{DG}{DF} \cdot \frac{FH}{GI}.$$ Combining the previous two results, we get: $$\frac{HK}{KI} \cdot \frac{HJ}{JI} = \frac{AC^2}{AB^2} \cdot \frac{DH \cdot FH}{DI \cdot IG}.$$ So it suffices to show that: $$\frac{DH \cdot HF}{DI \cdot IG} = \frac{AB^2}{AC^2}.$$ Lemma. $IHFG$ is cyclic. Proof. Invert at $D$ with arbitrary radius. We denote the inverses of points as usual. We have that $B', F', A'$ and $A', G', C'$ are collinear. We see that $\angle DB'A' = \angle DAB = \angle DAC = \angle DC'A'$, and so $A'B' = A'C'.$ We have $\frac{A'F'}{A'D} = \frac{AF}{DF} = \frac{AG}{GD} = \frac{A'G'}{A'D},$ so $A'F' = A'G'.$ Since $HI || BC$, we have that $(\triangle DH'I')$ is tangent to $BC.$ This means that $\angle DH'I' = \angle I'DC' = \angle G'DC' = \angle DG'F'$, which easily implies that $F'G'I'H'$ is cyclic (because $D = F'H' \cap G'I'.$ Inverting back implies the lemma. $\blacksquare$ From the lemma, we have that $\frac{DH \cdot HF}{GI \cdot ID} = \frac{DF' \cdot F'H'}{DG' \cdot G'I'}.$ Set $X = C'F' \cap B'G'$ and $Y = F'G' \cap DX.$ Since $B'F'G'C', DF'J'C',$ and $DB'J'G'$ are all cyclic (the first is an isosceles trapezoid, the others are images of lines under inversion), Radical Axis Theorem implies that $X \in DJ'.$ Note that $\angle DF'Y = \angle F'DB' = \angle H'J'D$, which implies that $F'YJ'H'$ is cyclic because $Y \in DJ'.$ This implies that $H'F' \cdot F'D = DF' \cdot DH' - DF'^2 = DY \cdot DJ' - DF'^2.$ Analogously, we have $G'I' \cdot G'D = DY \cdot DJ' - DG'^2.$ We have that $DY \cdot DJ' = DX \cdot DJ' \cdot \frac{F'C'}{XC'} = (DX^2 + DX \cdot XJ') \cdot \frac{F'C'}{XC'} = (DX^2 + F'X \cdot XC') \cdot \frac{F'C'}{XC'}.$ Hence, Stewart in $\triangle DF'C'$ with cevian $DX$ implies: $$DY \cdot DJ' - DF'^2 = \frac{C'D^2 \cdot F'X}{XC'}.$$ Analogously, $DY \cdot DJ' - DG'^2 = \frac{B'D^2 \cdot G'X}{XB'}.$ Combining the previous results yields that: $$\frac{DH \cdot HF}{DI \cdot IG}\frac{H'F' \cdot F'D}{G'I' \cdot G'D} = \frac{C'D^2}{B'D^2} = \frac{AB^2}{AC^2}.$$ From the above discussion, this finishes the problem. $\square$

Let $\Gamma$ be the circle tangent to $PQ$ and $PR$ at $Q$ and $R$ respectively. Let $RC$, $RD$, $QD$, $QB$ intersect $\Gamma$ at $X,Y,X’,Y’$. Note $XY\parallel CD\parallel DB\parallel X’Y’$ by homothety, and $\angle XRY=\angle CRD=\angle DQB=\angle X’QY’=\angle X’RY’$, so $X=X’$ and $Y=Y’$. Let $BX\cap CY=M$, and $KD\cap XY=N$. By Pascals, $KDMN$ collinear. Then $\frac{EK}{KF}=\frac{XN}{YN}=\frac{BD}{DC}$. By Reim, $BCQR$ concyclic. By radical center, let $AD$, $QB$, $RC$ concur at $T$. Also let $AD\cap XY=U$. Then, $\frac{FL}{LE}=\frac{YU}{UX}=\frac{BD}{DC}$. Since $\frac{EK}{KF}=\frac{FL}{LE}$, we have $EL=KF$

RaduAndreiLecoiu wrote: Claim $QRCB$ is cyclic Proof. This is just angle chasing. Let $QB \cap RD = \{S\}$ and $QD \cap RC = \{T\} $ Using that $AD$ is the bisector of angle $BAC$ then $\angle BQD = \angle DRC$ so $QRTS$ is cyclic. Since $QRCB$ is cyclic then by Reim's Theorem $BC \parallel ST$ . By Radical Axis $AD , CR, BQ$ are concurrent in $X$ Let $AD$ intersect $ST$ in $U$. Then $$\frac{EL}{LF} = \frac{TU}{US} = \frac{CD}{DB}$$Note that $\bigtriangleup RBS \sim \bigtriangleup QCT$ having 2 pairs of angles equal so the same holds also for the third hence $\angle CQD = \angle BRD$ From Law of Sines in $\bigtriangleup QKE$ and $\bigtriangleup RKF$ we get using the above equality $$\frac{EK}{KF} = \frac{QK}{RK} \cdot \frac{\sin{\angle RFK}}{\sin{\angle QEK}}$$But $\angle RFK = 180^{\circ} - \angle RDC$ and $\angle QEK = 180^{\circ} - \angle QDB$ so $$\frac{\sin{\angle RFK}}{\sin{\angle QEK}} = \frac{\sin{\angle RDC}}{\sin{\angle QDB}} = \frac{\frac{RC \cdot \sin{\angle DRC}}{DC}}{\frac{QB \cdot \sin{\angle BQD}}{BD}} = \frac{BD}{DC} \cdot \frac{RC}{BQ} = \frac{BD}{DC} \cdot \frac{RK}{QK}$$So $\frac{EK}{KF} = \frac {BD}{DC} = \frac{LF}{EL} $ Since $AB > AC$ the order of the points on the line parallel to $BC$ is $E,L,K,F$. Then in particular $EK + KF = EF$ and $LF + EL = EF$ so $\frac{EK}{EF} = \frac{LF}{EF}$ $\Rightarrow$ $EK = LF$. How to use angle chasing to prove $QRCB$ is cyclic?

Replace the labels $P, Q, R, K, E, L, F$ with $E, F, G, J, H, K, I$ respectively. So we now want $HK = JI$. We first prove $BCGF$ cyclic. Note that $EF^2=EA\cdot ED=EG^2 \Rightarrow EF=EG$. $\angle FBD=180^\circ - \angle FGC=\angle EGF+\angle GDC=\frac{1}{2}(180^\circ-\angle FEG)+\angle GDC$ $\Leftrightarrow 2\angle FBD=180^\circ-(360^\circ-\angle EFD-\angle EGD- \angle FDG)+2\angle GDC$ $\Leftrightarrow 2\angle FAE=-180^\circ+\angle EAF+\angle EAG+\angle FDG+2\angle GDC$ $\Leftrightarrow \angle FAE=-\angle FDB +\angle EAG+\angle GDC$ $\Leftrightarrow \angle EAB=\angle EAC$. Let $BF, DA, GC$ be concurrent at $P$. Let $HI$ intersect $BF$ and $CG$ at $V$ and $W$ respectively. Since $VW\parallel BC$, $$\frac{VH}{HJ}=\frac{VK}{KW}=\frac{JI}{IW}=\frac{BD}{DC}$$$$\frac{VJ}{HJ}=\frac{VW}{KW}=\frac{JW}{IW}=\frac{VW-JW}{KW-IW}=\frac{VJ}{KI}$$$$HJ=KI$$$$HK=JI$$

I believe the complete the complete relabelling of points used by Pathological and hxh123 isn't a coincidence, a source of the question paper did use those points, with a diagram attached as well.

That's why they replaced the label, very confusing at first...

[asy][asy] size(10cm); pair O = origin; pair A = dir(50); pair B = dir(200); pair C = dir(340); draw(A--B--C--cycle); pair D = extension(B, C, A, dir(270)); draw(A--D); pair O1 = circumcenter(A, B, D); pair O2 = circumcenter(A,D,C); filldraw(CR(O1, abs(D-O1)), invisible, lightred); filldraw(CR(O2, abs(D-O2)), invisible, lightred); pair P = 1.8*A-0.8*D; draw(P--A); pair Q = tangent(P, O1, abs(D-O1), 1); pair R = tangent(P, O2, abs(D-O2), 2); draw(P--Q); draw(P--R); pair S = extension(O1, O2, B, C); draw(C--S); draw(Q--R--S, lightblue); draw(O1--O2--S, lightblue); pair M = (B+D)/2; pair K = extension(Q, C, B, R); pair Kprime = extension(A, K, B, C); pair L = abs(K-A)/abs(Kprime-A) * (D-A)+A; pair E = extension(L, K, Q, D); pair F = extension(L, K, R, D); pair Qprime = abs(S-O2)/abs(S-O1) * (Q-S)+S; draw(E--F, yellow); draw(Q--C, lightgreen); draw(B--R, lightgreen); draw(Q--D, deepgreen); draw(R--D, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$O_1$", O1, dir(O1)); dot("$O_2$", O2, dir(O2)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$S$", S, dir(S)); dot("$M$", M, dir(M)); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$Q'$", Qprime, dir(Qprime)); [/asy][/asy] Let the circumcenters of $\bigtriangleup ABD$ and $\bigtriangleup ACD$ be $O_1$ and $O_2$, and let the respective circumradii be $r_1$ and $r_2$. Furthermore, let the distances of $O_1$ and $O_2$ from $BC$ be $d_1$ and $d_2$. Letting $O_1O_2$ intersect $BC$ at $S$, and setting the midpoint of $BD$ to be $M$, we obtain $$d_1 = O_1 M = BO_1 \sin\angle MO_1B = r_1 \sin\angle BAD$$Similarly, $d_2=r_2 \sin\angle CAD$. Noting that $\angle BAD=\angle CAD$, we obtain $\frac{d_1}{d_2}=\frac{r_1}{r_2}$. Hence $\frac{SO_1}{SO_2}=\frac{r_1}{r_2}$. Thus, $S$ is the exsimilicenter of the two circles $(ABD)$ and $(ACD)$. Let $QR$ intersect $(ACD)$ at another point $Q'$. By Power of a Point, $$PQ^2 = PA \cdot PD = PR^2$$which implies $PQ = PR$. Hence, $$\angle O_2 Q' R = \angle O_2 R Q' = 90^{\circ} - \angle PRQ = 90^{\circ}-\angle PQR = \angle O_1 QR$$implying that $O_1 Q \parallel O_2 Q'$, and thus $\frac{O_1 Q}{O_2 Q'}=\frac{r_1}{r_2}=\frac{O_1 S}{O_2 S}$. It follows that $Q$, $Q'$ and $S$ are collinear. Hence, $Q$, $R$ and $S$ are collinear. Notice that $$SQ\cdot SR = \frac{r_1}{r_2} SQ' \cdot SR = \frac{r_1}{r_2} SC\cdot SD = SD^2$$Thus $\bigtriangleup SQD \sim \bigtriangleup SDR$. This implies $$\angle DFE = \angle RDS = \angle DQS = \angle DQR$$and $\bigtriangleup DEF \sim \bigtriangleup DRQ$. We claim $\frac{EL}{FL}=\frac{r_2}{r_1}$. Observe that $$\frac{EL}{FL}=\frac{DE}{DF}\cdot \frac{\sin\angle QDA}{\sin \angle RDA} = \frac{DR}{DQ}\cdot \frac{\frac{AQ}{2r_1}}{\frac{AR}{2r_2}}=\frac{DR}{DQ}\cdot \frac{AQ}{AR}\cdot \frac{r_2}{r_1}$$Since $\bigtriangleup PAQ\sim\bigtriangleup PQD$, $\bigtriangleup PAR\sim \bigtriangleup PRD$, hence $$\frac{AQ}{QD}=\frac{PA}{PQ}=\frac{PA}{PR}=\frac{AR}{RD}$$This means that $\frac{EL}{FL}=\frac{r_2}{r_1}$. Since $\angle BQD = \angle BAD = \angle CAD = \angle CRD$, hence $$\angle BQS = \angle BQD + \angle DQS = \angle CRD + \angle SDR = \angle RCS$$which means that $B$, $C$, $R$ and $Q$ are concyclic. Furthermore, $$\angle BRD=\angle BRC-\angle CRD=\angle BQC-\angle BQD=\angle CQD$$Hence, $$\frac{EK}{FK}=\frac{\frac{EK}{\sin\angle EQK}}{\frac{FK}{\sin\angle FRK}}=\frac{QK}{RK}\frac{\sin\angle RFK}{\sin\angle QEK}=\frac{BQ}{CR}\frac{\sin\angle RDC}{\sin\angle QDB}=\frac{r_1}{r_2}$$Since $\frac{EK}{FK}=\frac{FL}{EL}$, we get $EL=FK$.

Let $M$ be the midpoint of arc $BC$ opposite $A$ on the circumcircle of $\triangle ABC$. Then $\measuredangle MBD=\measuredangle MBC=\measuredangle MAC=\measuredangle BAM$, so $\overline{MB}$ is tangent to $(ABD)$, and similarly $\overline{MC}$ is tangent to $(ACD)$. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=royalblue; pen sec=deepgreen; pen tri=deepcyan; pen qua=heavygreen; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pair A,B,C,M,D,T,P,Q,Qp,R,Rp,K,L,EE,F; A=dir(130); B=dir(210); C=dir(330); M=dir(270); D=extension(A,M,B,C); T=extension(B,C,(A+D)/2,(A+D)/2+rotate(90)*(A-D)); P=2A-D; pair[] qqq=intersectionpoints(circumcircle(A,B,D),circle( (P+circumcenter(A,B,D))/2,length(P-circumcenter(A,B,D))/2)); for (pair qq : qqq) { if (intersectionpoints(T--qq,P--M).length==0) { Q=qq; break; } } Qp=2*foot(circumcenter(A,B,D),T,Q)-Q; pair[] rrr=intersectionpoints(circumcircle(A,C,D),circle( (P+circumcenter(A,C,D))/2,length(P-circumcenter(A,C,D))/2)); for (pair rr : rrr) { if (intersectionpoints(T--rr,P--M).length>0) { R=rr; break; } } Rp=2*foot(circumcenter(A,C,D),T,R)-R; K=extension(B,R,C,Q); L=extension(A,D,K,K+C-B); EE=extension(Q,D,K,L); F=extension(R,D,K,L); filldraw(circumcircle(B,C,Q),tfil,tri); filldraw(circumcircle(A,B,D),sfil,sec); filldraw(circumcircle(A,C,D),sfil,sec); filldraw(circumcircle(A,B,C),fil,pri); fill(A--B--C--cycle,fil); draw(EE--F,qua+dashed); draw(Q--D--R,qua); draw(Q--P--R,tri); draw(B--M--C,tri); draw(P--M,tri); draw(B--R,sec); draw(C--Q,sec); draw(R--T--C--A--B,pri); dot("$A$",A,dir(150)); dot("$B$",B,SW); dot("$C$",C,SE); dot("$M$",M,S); dot("$D$",D,dir(250)); dot("$T$",T,W); dot("$P$",P,N); dot("$Q$",Q,dir(160)); dot("$Q'$",Qp,N); dot("$R$",R,dir(80)); dot("$R'$",Rp,SE); dot("$K$",K,N); dot("$L$",L,NE); dot("$E$",EE,dir(240)); dot("$F$",F,dir(300)); [/asy][/asy] The key claim is as follows. We present two different but instructive proofs. Claim. Points $B$, $C$, $Q$, $R$ are concyclic. First proof, by Pitot's theorem. Let $I=\overline{PQ}\cap\overline{MB}$ and $J=\overline{PR}\cap\overline{MC}$. Note that since $P$ and $M$ lie on the radical axis of $(ABD)$ and $(ACD)$, $PQ=PR$ and $MB=MC$. Furthermore by equal tangents, $IB=IQ$ and $JC=JR$. Summing, $PI+BJ=PJ+BI$, so by the converse Pitot's theorem, $PIMJ$ has an inscribed circle. Say it touches $\overline{MI}$, $\overline{IP}$, $\overline{PJ}$, $\overline{JM}$ at $B'$, $Q'$, $R'$, $C'$ respectively. Since $MB=MC$ and $MB'=MC'$, we have $\overline{BC}\parallel\overline{B'C'}$, and similarly all the sides of $BQRC$ and parallel to those of $B'Q'R'C'$. Since $B'Q'R'C'$ is cyclic, $\measuredangle BQR=\measuredangle B'Q'R'=\measuredangle B'C'R'=\measuredangle BCR$, so $BQRC$ is cyclic, as desired. $\blacksquare$ Second proof, by homothety. Let $T$ be the exsimilicenter of $(ABD)$ and $(ACD)$. Since $\widehat{BD}$ and $\widehat{CD}$ have equal measure, $T$ lies on line $BC$. We show that as $P$ varies on line $AD$, line $QR$ passes through $T$, and $TQ\cdot TR$ is fixed. Redefine $\overline{QR}$ as a line through $T$, intersecting $(ABD)$ and $(ACD)$ at $Q$ and $R$ on different sides of $\overline{AD}$. The task is to show that $P=\overline{QQ}\cap\overline{RR}$ lies on $\overline{AD}$, and $TQ\cdot TR$ is fixed. Let line $QR$ intersect $(ABD)$ and $(ACD)$ again at $Q'$ and $R'$, so that the homothety at $T$ sending $(ABD)$ to $(ACD)$ sends $Q$ to $R'$ and $Q'$ to $R$. Then if $P'=\overline{QQ}\cap\overline{QQ'}$, we have $P'Q=P'Q'$. By homothety, $\overline{Q'Q'}\parallel\overline{RR}$, so $\triangle P'QQ'\sim\triangle PQR$ and $PQ=PR$. It follows that $P$ has equal power fo $(ABD)$ and $(ACD)$, so $P\in\overline{AD}$. Finally, $TQ\cdot TR\propto TQ\cdot TQ'$, which is fixed by power of a point. $\blacksquare$ By Radical Axis theorem on $(ABD)$, $(ACD)$, $(BQRC)$, lines $AD$, $BQ$, $CR$ concur at a point $S$. Apply DDIT to $BQCR$ from $D$: we have that $(\overline{DB},\overline{DC})$, $(\overline{DQ},\overline{DR})$, $(\overline{DK},\overline{DS})$ are reciprocal pairs of some involution at $D$. Projecting onto $\overline{EKLF}$, we find that $(\infty_{EF},\infty_{EF})$, $(E,F)$, $(K,L)$ are reciprocal pairs of an involution on $\overline{EKLF}$, so it must be a reflection across some point on $\overline{EKLF}$. It follows that $\overline{EF}$ and $\overline{KL}$ share a midpoint, and we are done.

There is a very quick elementary finish after getting that $BCRQ$ is cyclic- Let $QD$ and $RD$ hit the circumcircle of $BCRQ$ for the second time at $Q'$ and $R',$ respectively. Clearly, $Q'R' \mid \mid BC$ and $X=CQ' \cap BR'$ lies on $AD.$ By Pascal on $CQQ'BRR'$ we ave that $Y=CR' \cap BQ'$ lies on $DK.$ Hence, it is enough to show that $Q'M=R'N,$ where $M=DX\cap Q'R'$ and $N=DY\cap Q'R',$ which is trivial.

Inverting at $D$ denote the inverse of points with primes. $\triangle ABC \mapsto \triangle A'B'C'$ with $A'B'=A'C'$. Claim: $P'Q' \parallel B'C'$ $P'$ is on line $A'D'$ and the condition that $PQ$ is tangent to $\odot ABD$ becomes $\odot DP'Q$ is tangent to line $AB$. Hence: $$A'Q^2=A'D \cdot AP'=A'R^2$$so $A'Q=A'R$ and assuming $Q,R$ lie on different sides of $AD$ this means $Q'R' \parallel B'C'$. Claim: $D,E,K'$ colinear where $S=B'R' \cap C'Q'$. Line $DK'$ radical axis of $\odot B'DR'$, $\odot C'DQ'$ but by symmetry: $$B'S \cdot SR'=C'S \cdot SQ'$$so $S$ lies on this line. Let $T,M$ be the midpoints of $B'C',Q'R'$ respectively. Then $A',M,E,T$ are colinear and by a well-known property of quadrilaterals $(A',E;M,T)=-1$. Let $DM \cap EF=N$ then: $$(E,F;N,\infty_{BC})=(DQ,DR;DN,D\infty_{BC})=(Q',R';M,\infty_{BC})=-1$$Hence $N$ is the midpoint of $EF$. Also as $D,E,K,K'$ colinear: $$(K,L;N,\infty_{BC})=(DK,DL;DN,\infty_{BC})=(E,A';M,T)=-1$$so $N$ is also the midpoint of $KL$. Hence using directed lengths: $$EL=EN+NL=NF+KN=KF$$

#3.... Quote: Let $\gamma_1,\gamma_2$ be two circles through $A,P$ such that $\gamma_1,\gamma_2$ are tangent to $MB,MC$ at $Q,R$ respectively can someone explain how? And what is DDIT

..bumppp

Dr_Vex wrote: #3.... Quote: Let $\gamma_1,\gamma_2$ be two circles through $A,P$ such that $\gamma_1,\gamma_2$ are tangent to $MB,MC$ at $Q,R$ respectively can someone explain how? And what is DDIT DDIT stands for dual of desargues involution theorem And for those 2 circles, we have 3 conditions so each circle is unique and easy to see those circles exist

redatced

Henry_2001 wrote: In triangle $ABC$, $AB>AC.$ The bisector of $\angle BAC$ meets $BC$ at $D.$ $P$ is on line $DA,$ such that $A$ lies between $P$ and $D$. $PQ$ is tangent to $\odot(ABD)$ at $Q.$ $PR$ is tangent to $\odot(ACD)$ at $R.$ $CQ$ meets $BR$ at $K.$ The line parallel to $BC$ and passing through $K$ meets $QD,AD,RD$ at $E,L,F,$ respectively. Prove that $EL=KF.$ Very nice problem and $Q,R$ being on opposite sides of $AD$ is required Solution- Claim- $BCQR$ is cyclic

It is enough to show that $\frac{EL}{LF}=\frac{FK}{KE} \iff (E,F,L,\infty_{BC})=(F,E,K,\infty_{BC})\iff (DQ,DR,DA,DC)=(DR,DQ,DK,DC)$ Now invert in $D$ and in the new figure - We get that $AB=AC$ and by claim we have $B,C,Q,R$ are concyclic $\implies QR\parallel BC$ Note that $\square BDRK,\square CDQK,\square BCRQ$ are cyclic and hence by radical axis theorem $DK,BR,CQ$ are concurrent at a point $T$ Let $DA\cap QR=Y, DK\cap QR=X, J=\infty_{BC}$ now we just do cross-ratio chase to finish the problem. $(DQ,DR,DA,DC)=(Q,R,Y,J)=(B,C,D,J)=(R,Q,X,J)=(DR,DQ,DK,DC)$ Hence proved !!!

Here's a different Solution. We Generalize the problem after proving $BCQX$ is a cyclic Quadrilateral rather than proceeding with Dual of Desargues Involution Theorem which was my first solution. Henry_2001 wrote: In triangle $ABC$, $AB>AC.$ The bisector of $\angle BAC$ meets $BC$ at $D.$ $P$ is on line $DA,$ such that $A$ lies between $P$ and $D$. $PQ$ is tangent to $\odot(ABD)$ at $Q.$ $PR$ is tangent to $\odot(ACD)$ at $R.$ $CQ$ meets $BR$ at $K.$ The line parallel to $BC$ and passing through $K$ meets $QD,AD,RD$ at $E,L,F,$ respectively. Prove that $EL=KF.$ We begin with a Lemma $\textbf{LEMMA :-}$ $ABC$ be a triangle with bisector $\overline{AD}$ and let $P$ be an arbitary Point on $\overline{AD}$. Let $\overline{PX},\overline{PY}$ be tangents to $\odot(ABD),\odot(ACD)$ where $X,Y\in\odot(ABD),\odot(ACD)$ respectively. Then $BCXY$ is a cyclic quadrilateral. Let $\Psi$ denote the composition of $\sqrt{AB\cdot AC}$ intersion around $A$ with a reflection around the angle bisector of $\angle BAC$. It suffices to prove the following property $\textbf{INVERTED PROPERTY:-}$ $ABC$ be a triangle and $M$ be the midpoint of arc $BC$ not containing $A$. Let $P$ be an arbitary point on $\overline{AM}$ and let $X,Y\in\overline{MB},\overline{MC}$ respectively such that $\odot(APX)$ and $\odot(APY)$ are tangent to $\overline{MB},\overline{MC}$ respectively then $BCXY$ is a Cyclic quad. Clearly $MX^2=MP\cdot MA=MY^2\implies MX=MY$ also $MB=MC$. So, $BCXY$ is an Isosceles trapezoid and hence a cyclic quadrilateral. So coming back to the Problem at hand we have $BCQR$ is a cyclic quadrilateral. So, by Radical Axis Theorem on $\odot(BCQR),\odot(ADB),\odot(ACD)$ we have $\overline{BQ},\overline{AD},\overline{CR}$ concurrent. Now we totally forget the angle bisector condition that is we treat $D$ as an arbitary point on $\overline{BC}$ and prove the following generalization. $\textbf{GENERALIZATION:-}$ Let $ABCD$ be a cyclic quadrilateral and let $\overline{AC}\cap\overline{BD}=\{P\}$.Let $\ell$ be the line through $P$ parallel to $CD$ and $X$ be an arbitary point on $\overline{CD}$ and $\tau$ be the line through $X$ such that $\overline{AD},\overline{BC},\tau$ are concurrent. Let $\ell\cap\tau=\{K\}$. $\{\overline{XA},\overline{XB}\}\cap\ell=\{U,V\}$. Then $UP=KV$. Animate $\{X\}$ on $\overline{CD}$. Then $X\mapsto U\mapsto K\mapsto V$ are Homographies on $\ell$. Let $M$ be the midpoint of $UV$. Then $(U,V;M,\infty_{\ell})=-1$ and as $U,V$ move projectively on $\ell$ so $M$ will also move projectively on $\ell$ and if $M^*$ is the midpoint of $PK$ then $X\mapsto K\mapsto M^*$ will be a Homography on $\ell$ as it is a Homothety at $P$ with a fixed ratio of $0.5$. So, $M^*\equiv M$ on three input values of $X$ on $\overline{CD}$ and clearly $X=C,D$ are trivial cases and when $X\equiv \tau\cap\overline{CD}$ then $(A,B;\tau\cap\overline{AB},\overline{AB}\cap\overline{CD})\overset{X}{=}(U,V;P,\infty_{\ell})\implies PU=PV$ as desired. $\blacksquare$

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We have $$PQ^2=PA\times PD=PR^2$$$\blacksquare$ CLAIM 2. $QB,AD,RC$ are concurrent Proof. Applying Ceva's theorem to $\triangle ABC$, it suffices to show $$\frac{\sin\angle QBA}{\sin\angle QBD}\cdot\frac{\sin\angle RCB}{\sin\angle ACR}=1$$Which is true since $$\frac{\sin\angle QBA}{\sin\angle QBD}=\frac{\sin\angle PQA}{\sin\angle PAQ}=\frac{PA}{PQ}=\frac{PA}{PR}=\frac{\sin\angle ACR}{\sin\angle RCB}$$$\blacksquare$ CLAIM 3. Suppose $QB,AD,RC$ are current at $T$ then $Q,A,R,T$ are concyclic. Proof. $$\angle TQA=\angle BQA=\angle ADC=180^{\circ}-\angle ARC=180^{\circ}-\angle ART$$$\blacksquare$ CLAIM 4. $EL\cdot EF=FK\cdot FL$ Proof. $$\frac{EK}{KF}=\frac{DC\cdot\frac{QK}{QC}}{DB\cdot\frac{RK}{RB}}=\frac{DC}{CB}\cdot\frac{QK}{RK}\cdot\frac{QC}{RB}$$Meanwhile, by easy angle chasing we have $\triangle DLE\sim \triangle BQA$ and $\triangle LDF\sim\triangle RCA$. Hence $$\frac{EL}{LF}=\frac{EL}{LD}\cdot\frac{LD}{LF}=\frac{QA}{QB}\cdot\frac{RC}{RA}=\frac{QA}{RA}\cdot\frac{KC}{KB}$$Notice that $\frac{QK\cdot KC}{RK\cdot KB}=1$, hence it suffices to show $$\frac{AC}{AB}\cdot\frac{QC}{RB}\cdot\frac{QA}{RA}=1$$Notice that the Left Hand Side equals $$\frac{AC}{AB}\cdot\frac{TC}{TB}\cdot\frac{\sin\angle DRC}{\sin\angle DRB}=\frac{AC}{AB}\cdot\frac{BD}{DC}=1$$so we are done. $\blacksquare$ This obviously implies $FK=EL$ so we are done.

Nice problem! First let: $S \equiv RD \cap CQ$ , $T \equiv BR \cap DQ$ , $W \equiv QD \cap RC$ , $X \equiv RD \cap BQ$ , $G \equiv AD \cap (ABC)$ Now as mentioned above we have $RQBC$ is cyclic is lemma since $GB, GC$ are tangent to $(ABD), (ACD)$ respectively. Claim 1: $RQTS$ is cyclic and $ST || BC$ Proof: Note that since $RQBC$ is cyclic we have $\angle CRB= \angle CQB$ and since $\angle CRD=\angle DQB=\frac{A}{2}$ we indeed get $\angle SRT=\angle SQT$ and then $\frac{KS}{KT}=\frac{KR}{KQ}=\frac{KC}{KB}$ by POP of $K$, $\implies$ $ST || BC$ Proving the claim. Claim 2: $BE , CF , KD$ concur on $QR$ proof: Since $FE || BC$ we want to prove that $\frac{KF}{KE}=\frac{CD}{BD}$ and the claim follows by the inverse of Pappus threorem on $(B,D,C),(Q,P',R)$ where $P' \equiv BE \cap CF$ Note that by Thales: $\frac{KF}{ST}=\frac{RK}{RT}$ , $\frac{KE}{ST}=\frac{KQ}{QS}$ $\implies$ $$\frac{KF}{KE}=\frac{RK}{QK}.\frac{QS}{RT}=\frac{KS}{KT}.\frac{DS}{DT}=\frac{KC}{KB}.\frac{sin\angle CKD}{sin \angle BKD}=\frac{CD}{BD} $$(Since $\angle KTD=180-\angle DRT-\angle RDT=180-\angle DQS-\angle QDS=\angle KSD$, which implies: $\frac{DS}{DT}=\frac{sin\angle CKD}{sin\angle BKD}$) Claim 3: $\frac{LE}{LF}$=$\frac{CD}{BD}$ Proof: We can use Pappus again to prove that $O \equiv BF \cap CE $ is in fact $AD \cap ST$ which implies the claim since $\triangle COB \sim \triangle FOB$. Or just note that $\angle RXB+\angle DRB=\angle RBQ=\angle RCQ=\angle QWC+\angle CQT$ $\implies$ $RQXW$ is cyclic, Thus $\angle DWX=\angle XRQ=\angle STW$ which implies that $\triangle TSD \sim \triangle WXD$. Now since $AD,CR,BQ$ concur, the claim follows. Finally since $\frac{FK}{EK}=\frac{LE}{LF}$ we have $KF=LE$.