Let $M$ be the midpoint of $BC$ and let $K_A,L_A$ be the feet of the internal and external angle bisectors of $\angle BAC$ on $BC$. Let $H$ be the orthocenter of $\triangle ABC$ and let $A_1$ be the reflection of $A$ over the midpoint of $BC$. First, by angle chasing, since $\angle KBC+\angle KCB=\angle KAB+\angle KAC= \angle A$, we know that $\angle BKC=180^{\circ}-\angle A = \angle BFC$, hence $A,B,F,C$ are concyclic. It's also clear that $AK$ bisects $BC$ by Power of a Point. Since $BHKC$ is cyclic we know $\angle HKA_1=90^{\circ} \implies \angle HKA=90^{\circ}$, so $APQHK$ is cyclic. We know that $MK\cdot MA = MB^2$. Since $(B,C;K_A,L_A)$ is harmonic, it's well-known that $MB^2=MK_A\cdot ML_A$, hence $MK_A\cdot ML_A = MK\cdot MA$ implying that $AKK_AL_A$ is cyclic on the $A$-Apollonius circle of $\triangle ABC$. This implies that $KB:KC = FB:FC = AB:AC$, hence $(A,F;B,C)$ is harmonic. Now notice that $AF$ is the $A$-symmedian of $\triangle ABC$, but since $\triangle APQ,\triangle ABC$ are inversely similar in $\angle BAC$, $AF$ must be the median of $\triangle APQ$. Hence it suffices to show that $E$ is on the median of $\triangle APQ$. By angle-chasing, $\angle EQP=\angle KQP=\angle KAP=\angle KCB$, and similarly $\angle EPQ=\angle KBC$, implying that $E,K$ are corresponding points with respect to similar triangles $\triangle APQ,\triangle ABC$, so $AE$ must bisect $PQ$ as desired.

From Geometry!, $AF$ is the $A$-symmedian of $\triangle ABC$. Let $M$ be the midpoint of $BC$; then $MB^2=MC^2$ so $M$ lies on the radical axis $AK$ of $C_1, C_2$. Since $M$ is also the center of $(BQPC)$, then $MA\cdot MK = MP^2\implies MP$ is tangent to $(APK)$. However, since $MP$ is tangent to $(APQ)$, then $A, P, K, Q$ are concyclic. In particular, since $AK$ is the $A$-median of $\triangle ABC$, then $AK$ is the $A$-symmedian of $\triangle APQ$ so by the reverse of Geometry!, $AE$ is the $A$-median of $\triangle APQ$ so $AE$ is the $A$-symmedian of $\triangle ABC$. But $AF$ is the $A$-symmedian already so $A, E, F$ collinear, done.

It is well-known that the point symmetric to $K$ in the line $BC$ lies on the $A$ symmedian in triangle $ABC$. Let $H$ be the orthocenter of triangle $ABC$. Note that the points $A,P,Q,K$ lie on a circle with diameter $AH$. As $AK$ is a median in triangle $ABC$, it is the $A$ symmedian in triangle $APQ$ (due to anti parallelism) and the point symmetric to $K$ in the line $PQ$ lies on the median from $A$ in triangle $APQ$, which is just the $A$ symmedian in triangle $ABC$. The points $A,E,F$ are thus, collinear.

As we can see in the previous posts, essentially the hardest part of the solution is to prove that quadrilateral $APKQ$ is cyclic. In fact, we can prove that $APHKQ$ is a cyclic pentagon, where $H$ is the orthocenter of $\bigtriangleup ABC$. Let $H'$ be the reflection of $H$ across $BC$. Therefore, $KH,\ FH'$ and $BC$ concur at a point $X$. Clearly, it's enough to show that quadrilateral $QKHP$ is cyclic, then, by the radical axis theorem, we only need to prove that $QP, KH, BC$ concur. Let $M$, as usual, be the midpoint of side $BC$. Consider the circumcircle of $BQHP$; we know $M$ is its circumcenter. Let $PQ\cap BC=X$. Notice that $HX$ is the polar of $A$ wrt circumcircle of $BQPC$, then, by La Hire's theorem, $K$ will lie on $HX$ iff $A$ lies on the polar of $X$ wrt circumcircle of $BQPC$. It's easy to derive that $M,K,A$ are collinear, and using that $BC$ is tangent to $C_1,C_2$ we find that $MA\cdot MK=MB^2$, hence the result follows immediately.

Another approach: Let $H$ be the orthocenter of $\triangle ABC$. Let $\angle CAK=\angle KCB = \alpha$ and $\angle BAK=\angle KBC = \beta $. Now we get that $\angle BKC= \angle BHC= 180^{\circ} - \alpha -\beta$, so $BCKH$ is cyclic, so $\angle PBK= \angle QCK$. Now we'll prove that $\triangle PBK\sim QCK$, for which we'll need to show that $\frac{BK}{CK}=\frac{QC}{BP}$. But we know that $\frac{QC}{BP}=\frac{AC}{BC}$. Now we'll perform an inversion around $A$ with some radius and let point $X$ go to point $X'$ for every point. Now $K'C'=K'B'$, because they are tangents to the same circle, so $\frac{r^2}{AK.AC}\cdot KC= \frac{r^2}{AK.AB}\cdot KB \Leftrightarrow AC\cdot BK=AB\cdot CK$, which proves the claim that $\triangle BKP\sim \triangle CKQ$. We get that $\angle BKC=\angle QKP=\angle QHP$ ,so $QPKH$ is cyclic, so $\angle QPK=\beta$ and $\angle PQK= \alpha$ so we get that the following figures are similar in this order $ APKQ\sim ABFC$ . We are done because of respective elements in similar figures(if that's what it's called) we get that $\angle QAE=\angle PAK = \angle BAF $, which proves that $A,E$ and $F$ are collinear.$\blacksquare$

It is clear that $K$ is the HM-point of $\triangle ABC$. Now we use the fact that the reflection of the HM-point of a triangle $ABC$ in $BC$ is the intersection of the $A$-symmedian with $(ABC)$ from Theorem 1.4.8.2 here to see that $AF$ is the $A$-symmedian of $\triangle ABC$ and $AE$ is the $A$-median of $\triangle APQ$. So since they are the same line $A, E$ and $F$ are collinear.

Overly long solution First, let $M$ and $N$ denote the midpoint of $AB$ and $PQ$, $H$ the orthocenter of $\triangle ABC $ and $H_1$ be the orthocenter of $\triangle AQP$. Notice that $A,K,M$ are collinear by radical axis Claim 1: $BHKC$ and $AQKP$ are cyclic Indeed, we have $\angle KAC = \angle KCM$ so, $\angle MKC = \angle C$ by similar triangles. Thus, $\angle BKC = \angle B + \angle C$ as desired. Now, notice that $\angle BKM = \angle B = \angle BQM$ implying $BQKM$ and $CMKP$ are cyclic. Thus, by Miquel, $AQPK$ is cyclic. $\square$ Claim 2: $A,N,F$ are collinear. First, notice that $(AHB)$ is the reflection of $(ABC)$ over $AB$. So, $F$ lies on $(ABC)$. We have \[ \angle FAC = \angle FBC = \angle CBK = \angle BAM \] Notice that $\angle BAM = \angle PAN$ since $\triangle ABC$ and $\triangle AQP$ are similar. Thus, $\angle FAC = \angle NAC$ as desired. $\square$. Claim 3: $A,N,E$ are collinear Notice that $A,K,M$ are collinear and $\triangle ABC$ and $\triangle AQP$ are similar. So, it is sufficient to show $E$ wrt $\triangle AQP$ is the same as $K$ wrt $\triangle ABC$. Indeed, notice that \[ \angle EPQ = \angle QPK = \angle CHK = \angle CBK \] Similarly, $\angle PQE = \angle KCB$. So, $\triangle QEP$ and $\triangle CKB$ are similar. Implying the claim. $\square$ Thus, $A,N,E,F$ are collinear as desired. $\blacksquare$

what is this question.

Just notice $K$ is the $A-Humpty$ Point of $ABC.$ Call $H$ the orthocenter. Now, $K$ so it lies on $(AHPQ)$ and $(BCH).$ As $PQ$ and $BC$ are antiparallels, then reflecting $K$ across $PQ$ gives a point of the median of $(APQ),$ and reflecting $K$ over $BC$ gives a point on the symeddian of $(ABC),$ implying that we are done.

?? Note that $K$ is the $A$-HM point of $\triangle ABC$, so it's well known that $F$ is just the intersection of the $A$-symmedian of $\triangle ABC$ with $(ABC)$. Additionally, $K$ lies on the $A$-median of $\triangle ABC$, so it lies on the $A$-symmedian of $\triangle APQ$ as $\overline{QP}$ and $\overline{BC}$ are antiparallel in $\angle A$. But $K$ also lies on $(APQ)$, so $E$ must in fact be the $A$-HM point of $\triangle APQ$, and therefore lies on the $A$-median of this triangle. Noting once again that $\overline{QP}$ and $\overline{BC}$ are antiparallel in $\angle A$, this means that $E$ lies on the $A$-symmedian of $\triangle ABC$, as desired.