This length bash was very clean and satisfying. I will give only a sketch. Firstly, let WLOG $\alpha$ be the smallest angle. Easy angle chasing gives $\angle PBK= \gamma - \alpha, \angle QCL = \beta - \alpha$. It is easy to see that proving $\angle PKB= \angle CQL$ is sufficient. By LoC for $\triangle PBK$ and with a bit of trig manipulations (using LoC in $\triangle ABC$ as well), one will obtain that $PK=\frac{c^2-a^2}{b}$ and similarly $QL=\frac{b^2-a^2}{c}$. Now, once again LoC in $\triangle PKB$ with a few manipulations gives that $\cos(\angle PKB)=\frac{b^2+c^2-a^2}{2bc}$, and since it is entirely symmetric, we are done (and particular, this proves that the $\angle PKB =\angle CQL = \alpha$).

Let $X, Y, Z$ be the midpoints of segments $KA, AB, BC$. Let $W$ be the midpoint of $PB$. Let $\alpha = \dfrac{\angle{BAC}}{2}$. Then $\angle{KBX} = \alpha = \angle{XBA}$ which implies that $KB \| XY$. But we already knew that $YZ \| AC$ so $X, Y, Z$ are collinear. Since $PB = BC$, then $WB = BZ$ and $\angle{BWZ} = \alpha = \angle{BZW}$ because $PB$ is tangent to $(ABC)$. Also, $\angle{BXA} = 90^{\circ}$ so $\angle{BXY} = \angle{XBY} = \alpha$. Therefore $\angle{BXZ} = \angle{BWZ}$ and we conclude that $B, W, X, Z$ are concyclic. Then $\angle{WXB} = \angle{WZB} = \alpha = \angle{XBA}$ so $WX \| BA$. It implies that $PK \| WX \| BA$. Analogously, $QL \| CA$. Finally, $\angle{PKA} = \angle{QLA}$ and we're done.

Let $AS$ be the angle bisector fo $\angle BAC$ such that $S$ lies on $\Gamma$ and $T$ be the topmost point of arc $\widehat{BAC}$. Let $\angle BAS=\alpha$, $\angle TAC=\beta$, $\angle ABC=\gamma$, $\angle ACB=\phi$. Claim: Quadrilateral $PBCQ$ is an isosceles trapezoid which is then cyclic with center $T$. Proof. Notice that $\angle PBC=\gamma+\phi=\angle QCB$ and $PB=BC=CQ$. And $T$ lies on the perpendicular bisector of $BC$ and $\angle BTC=2\alpha=\frac{1}{2}\angle BQC$. $\blacksquare$ Claim: Lines $AC\parallel QL$, similarly $AB\parallel BK$. Proof. Since $T$ is the center of cyclic quad $PBCQ$ therefore notice that $BCQT$ is a kite. Hence $\angle TLC=\beta=\angle TBC=\angle TQC$, gives $TLQC$ is cyclic. So $\angle CLQ=\angle CTQ=\angle BTC=\angle BAC=\angle CAL$, means $AC\parallel QL$. Similarly $AB\parallel BK$. $\blacksquare$ From the above claims we can say $\angle PKL=\angle QLK$ means $MK=ML$.

Easy to see that $BK\parallel AC$. Let $X = KB\cap \Gamma$. It is not dificult to show that $\triangle BXC = \triangle PKB$ this implies $KP\parallel AB$. Analogously $LQ\parallel AC$. The rest is angle chasing.

We can directly compute the angles $\angle QLA$ and $\angle AKP$. If we consider $Q'\neq B\in (ABC)$ such that $CB=CQ'$, then we can show $\triangle CAQ'=\triangle CLQ$. Hence, directing angles: $$\angle QLA=\angle QLC+\angle CLA=\angle Q'AC+\angle BAL=\angle CAB+\angle BAL$$and similarly $\angle AKP=\angle CAB+\angle KAP$. These values are equal. $\square$

This problem was proposed by me, Nuno Arala, from Portugal. I hope you enjoyed it! In case anyone is interested (and is able to read Portuguese), I am attaching a copy of the original submission.

Solution from Twitch Solves ISL: Let $Y$ be the arc midpoint of arc $BAC$. Claim: $P$ is reflection of $C$ across line $YB$. Proof. It's easy to check $\overline{BY}$ bisects $\angle PBC$. $\blacksquare$ Claim: $PKBY$ is cyclic. Proof. $\measuredangle BPY = \measuredangle YCB = \measuredangle YAB = \measuredangle KAB = \measuredangle BKA = \measuredangle BKY$. $\blacksquare$ Now $\measuredangle PKY = \measuredangle PBY = \measuredangle YCQ = \measuredangle YLQ$ so $\triangle MKL$ has isosceles base angles. [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 */ pair C = (1.5,-1.); pair B = (-5.30397,-0.96599); pair A = (-4.,8.); pair P = (-10.51689,3.40672); pair Q = (6.75636,3.32040); pair O = (-1.88151,3.11407); pair Y = (-1.85490,8.43944); pair K = (-10.02848,6.76501); pair L = (3.01801,9.43770); import graph; size(12cm); pen zzttqq = rgb(0.6,0.2,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(circle(O, 5.32543), linewidth(0.6) + blue); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(K--L, linewidth(0.6)); draw(P--(-1.91610,-3.80782), linewidth(0.6) + yqqqqq); draw(Q--(-1.91610,-3.80782), linewidth(0.6) + yqqqqq); draw(P--K, linewidth(0.6)); draw(Q--L, linewidth(0.6)); draw(circle((-5.27736,4.35936), 5.32543), linewidth(0.6) + linetype("2 2") + qqwuqq); draw(B--Y, linewidth(0.6) + blue); draw(Y--C, linewidth(0.6) + blue); draw(circle((1.52661,4.32536), 5.32543), linewidth(0.6) + linetype("2 2") + qqwuqq); dot("$C$", C, dir((10.982, -57.148))); dot("$B$", B, dir((-21.122, -47.876))); dot("$A$", A, dir((-50.135, 34.157))); dot("$P$", P, dir((-56.031, -82.430))); dot("$Q$", Q, dir((9.161, -48.452))); dot("$O$", O, dir((-50.768, -34.860))); dot("$Y$", Y, dir(95)); dot("$K$", K, dir((-48.548, 23.886))); dot("$L$", L, dir((5.623, 11.484))); [/asy][/asy] Now $\measuredangle PKY = \measuredangle PBY = \measuredangle YCQ = \measuredangle YLQ$ so $\triangle MKL$ has isosceles base angles. Remark: Conjectures to prove $(APQ)$ is tangent to $\overline{KL}$ and $M$ lies on the $A$-symmedian.

Conjuring! a_507_bc wrote: Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $P$ and $Q$ be points in the half plane defined by $BC$ containing $A$, such that $BP$ and $CQ$ are tangents to $\Gamma$ and $PB = BC = CQ$. Let $K$ and $L$ be points on the external bisector of the angle $\angle CAB$, such that $BK = BA, CL = CA$. Let $M$ be the intersection point of the lines $PK$ and $QL$. Prove that $MK=ML$. Note that $MK=ML$ will follow if we can show that $PK$ and $QL$ make equal angles with line $KL$ with opposite orientations. Since $\angle AKB=\angle ALC=90^{\circ}-\tfrac{1}{2}\angle A$, we only need to show that $\angle BKP=\angle CLQ$. These two angles are quite ``disconnected", but we see a natural rotation at $B$ taking $A$ to $K$ and at $C$ taking $L$ to $A$. So we conjure points $S$ and $T$ which are the images of $P$ and $Q$ in the above rotations respectively. So $BS=BC$ and $\angle SBC=\angle PBC-\angle PBS=180^{\circ}-\angle A-\angle KBA=180^{\circ}-2\angle A$ hence $\angle BSC=\angle BCS=\angle A$, so $S$ lies on $(ABC)$. Hence $\angle BKP=\angle BAS=\angle BCS=\angle A$. Similarly, $\angle CLQ = \angle A$. Thus, the conclusion follows.