Using ceva threom we obtain that BG=CG By angle chasing we acquire that ⊿BQD∽⊿BNC ⊿CQD∽⊿CMB => DH*BC=DC*BM=CN*BD Notice AB/AC=BM/CN => DC*AB=AC*BD ABCD is a harmonious quadrilateral This completes our proof

Here is my solution (Using spiral similarity, no calculation) Because $ (\overline {BMA},\overline {BPN},\overline {CMP},\overline {CNA})$ is a complete quadrilateral, therefore, $ Q$ is the Miquel point wrt $ (\overline {BMA},\overline {BPN},\overline {CMP},\overline {CNA})$. Therefore, $ (A,N,Q,B)$, and $ (A,M,Q,C)$ are the sets of concyclic points. Hence, $ \angle BAQ = \angle BNQ\equiv \angle PNQ = \angle QCP$. Therefore, we just need to prove that $ \angle QCP = \angle PAN$. Indeed, it is easy to notice that $ \triangle QAB\sim \triangle QCP$, further, these $ 2$ triangles have the same direction, hence, through the spiral similarity with center $ Q$, denote this $ f$, $ f: \triangle QAB\mapsto \triangle QCP$, therefore, $ f: A\mapsto C$, $ B\mapsto P$ $ \Longrightarrow f: A\mapsto B$, $ C\mapsto P$, which implies that $ f: \triangle QAC \mapsto \triangle QBP \Longrightarrow \frac {QC}{QP} = \frac {AC}{BP}$, $ (1)$. Now, by Thales theorem, we notice that: $ \frac {AC}{AN} = \frac {BC}{MN} = \frac {PB}{PN} \Longrightarrow \frac {AC}{PB} = \frac {AN}{PN}$, $ (2)$. From $ (1),(2)$, we obtain, $ \frac {NA}{NP} = \frac {QC}{QP}$, combine with $ \angle ANP = \angle CQP\Longrightarrow \triangle ANP\sim \triangle CQP\Longrightarrow \angle PAN = \angle PCQ$, which implies to the result of the problem. Our proof is completed

We have: $ \frac {AN\sin{\widehat{QAN}}}{AM\sin{\widehat{QAM}}} = \frac {S_{AQN}}{S_{AQM}} = \frac {AN}{AM}.\frac {QN}{QM}.\frac {\sin{\widehat{QNC}}}{\sin{\widehat{QMB}}}$ $ \Rightarrow \frac {\sin{\widehat{QAN}}}{\sin{\widehat{QAM}}} = \frac {QN}{QM}.\frac {\sin{\widehat{QNC}}}{\sin{\widehat{QMB}}} = \frac {QM}{QB}.\frac {QN}{QM} = \frac {QN}{QB} = \frac {AN}{AM} (1)$ It is because of the fact that $ \triangle QCN \sim \triangle QMB$ and since, $ \frac {QN}{QB} = \frac {AN}{AM}$. From $ (1)$, it is followed that $ AQ$ is the symmedian of $ \triangle AMN$, which implies the result of the problem. P/S: My $ 200^{\text{th}}$ post

Ahiles wrote: Let $ MN$ be a line parallel to the side $ BC$ of a triangle $ ABC$, with $ M$ on the side $ AB$ and $ N$ on the side $ AC$ . The lines $ BN$ and $ CM$ meet at point $ P$ . The circumcircles of triangles $ BMP$ and $ CNP$ meet at two distinct points $ P$ and $ Q$. Prove that $ \angle BAQ = \angle CAP$ . Proof. $ \left\|\begin{array}{c} \widehat {BMQ}\equiv\widehat {BPQ}\equiv\widehat {NCQ} \\ \ \widehat {MBQ}\equiv\widehat {CPQ}\equiv\widehat {CNQ}\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c}\widehat {BMQ}\equiv\widehat {NCQ} \\ \ \widehat {MBQ}\equiv\widehat {CNQ}\end{array}\right\|$ $ \implies$ $ \triangle BMQ\sim\triangle NCQ\implies$ $ \frac {AB}{AC} = \frac {BM}{NC} = \frac {\delta (Q,BM)}{\delta (Q,NC)}$ $ \implies$ $ \frac {\delta (Q,AB)}{\delta (Q,AC)} = \frac {AB}{AC}$ , i.e. the point $ Q$ belongs to the $ A$-symmedian. I used the notation $ \delta (X,d)$ - the distance of the point $ X$ to the line $ d$ . I"ll present below two simple and well-known properties : $ 1\blacktriangleright$ The geometrical locus of an interior point $ L$ of the angle $ \widehat {BAC}$ for which $ \frac {\delta (L,AB)}{\delta (L,AC)} = \frac {AC}{AB}$ is the $ A$-median. $ 2\blacktriangleright$ The geometrical locus of an interior point $ L$ of the angle $ \widehat {BAC}$ for which $ \frac {\delta (L,AB)}{\delta (L,AC)} = \frac {AB}{AC}$ is the $ A$-symmedian. Virgil Nicula wrote: An easy extension. Consider in $ \triangle ABC$ two mobile points point $ M$ , $ N$ which belong to the sidelines $ AB$ , $ AC$ respectively so that $ \frac {MB}{NC} = k$ (constant) and the line $ BC$ doesn't separates the points $ M$ , $ N$ . Denote $ P\in MC\cap NB$ and the second intersection $ Q$ between the circumcircles of the triangles $ BMP$ , $ CNP$ . Prove that $ \frac {\delta (Q,AB)}{\delta (Q,AC)} = k$ , i.e. the geometrical locus of the point $ L$ is a line which passes through $ A$ .

a Turkish student solve it from four trigonometric ceva

Let $ H$ be the midpoint of $ BC$. Applying the lemma about trapezium we have $ A,P,H$ are collinear then $ AP$ is the median of triangle $ ABC$. We need to prove $ AQ$ is a symmedian. Let $ K, J$ be the projection of $ Q$ on $ AB, AC$. $ \angle BMQ = \angle BPQ = \angle NCQ, \angle MBQ = \angle MPB = \angle NQC$ $ \Rightarrow \Delta MBQ\sim \Delta CNQ \Rightarrow \frac {AB}{AC} = \frac {BM}{CN} = \frac {QK}{QJ}$ $ \Rightarrow$ QED

Complete quadrilateral - need symmedian - spiral similarity - done: http://tlovering.wordpress.com/2009/05/05/balkan-mo-2009-q2-a-geometers-geometry/

Note that the problem was posted 2 years ago from the competition in that forum. See here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=175434

You can see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=358752 Sincerely Jean-Louis

I solved this using lots of ratios and sine rules. Let $AP$ meet $BC$ at $X$. Let $\angle BMQ = \angle ACQ = x$, $\angle QBA = \angle QNC = y$ and $\angle MQB = \angle NQC = z$. Here we make the observation that $Q$ is the Miquel Point of the compelte quadrilateral with vertices $B, M, A, N, C, P$. $\frac{AQ}{\sin x} = \frac{MQ}{\sin \angle BAQ}$ and $\frac{AQ}{\sin y} = \frac{NQ}{\sin \angle CAQ}$ so $\frac{\sin x}{\sin y} = \frac{\sin \angle BAQ}{\sin \angle CAQ} \cdot \frac{NQ}{MQ}$ (1). $\frac{NQ}{\sin x} = \frac{NC}{\sin z}$ and $\frac{MQ}{\sin y} = \frac{MB}{\sin z}$ so $\frac{NQ}{MQ} = \frac{\sin x}{\sin y} \cdot \frac{AC}{AB}$ (2). (1), (2) $\implies \frac{\sin \angle BAQ}{\sin \angle CAQ} = \frac{AB}{AC}$, and it is quite well known that this implies that $AQ$ is a symmedian of $\triangle ABC$. Using Ceva, we get $\frac{BM}{MA} \cdot \frac{AN}{NC} \cdot \frac{CX}{XB} = 1 \implies CX = BX$ as $MN || BC$. Thus $AX$ is a median, so $\angle BAQ = \angle CAQ$, as required.

Thus $AX$ is a median, so $\angle BAQ = \angle CAQ$, as required.[/quote] BAP=CAQ

A simple proof which highlights the configuration: We proceed using directed angle $\pmod{\pi}$. Let $AP\cap BC=X$, by Ceva $AP$ is a median of $\triangle AMN$. Now note that \[\measuredangle ABQ=\measuredangle MBQ=\measuredangle MPQ =-\measuredangle QPM =-\measuredangle QPC =\]\[=-\measuredangle QNC=-\measuredangle QNA =\measuredangle ANQ.\]Hence, $ABNQ$ is cyclic. Similarly, $ACMQ$ is cyclic. Let $F=(CNPQ)\cap MQ$. We claim that $NF\parallel AB$. Now $Q$ is the center of the unique spiral similarity sending $BM$ to $NC$. Then \[\measuredangle QMB =\measuredangle QCN =\measuredangle QFN =-\measuredangle NFQ =-\measuredangle NFM=\measuredangle MFN.\]So $NF\parallel AB$. Similarly, if $G=BMPQ\cap NQ$, then $MG\parallel AC$. Let $H$ be the point such that $AMHN$ is a parallelogram. Then $H$ lies on $MG,NF,AP$. Note that $\measuredangle PGN =\measuredangle PGQ=\measuredangle PMQ=\measuredangle PMF.$ Similarly, $\measuredangle PFM=\measuredangle PNG.$ So $P$ is the center of the unique spiral similarity sending $MF$ to $GN$, so $P=(MFH)\cap (NGH)$ and hence $MFHP,NGHP$ are cyclic. We are close! Note that $\measuredangle ANP =\measuredangle ANB=\measuredangle AQB$ and \[\measuredangle NPA=\measuredangle NPH =\measuredangle NGH=\measuredangle NGM=-\measuredangle MGN =-\measuredangle MGQ=\]\[=-\measuredangle MBQ=\measuredangle QBM=\measuredangle QBA.\]So $\triangle PAN\sim \triangle BAQ$, and $\angle PAN=\angle PAC=\angle BAQ$, as desired. $\square$

My solution : By Miquel's theorem $AMQC,BANQ$ are cyclic quads. Apply $\sqrt bc$ inversion with centre $A$. We get that 1.) $B$ goes to $C$, $C$ to $B$ 2.) Let $M$ go to $M'$ and $N$ go to $N'$ and $Q$ go to $Q'$ 3.)Since a circle through the centre of inversion becomes a line we get that $ Q' $ is the intersection of $BM'$ and $CN'$. So $AQ'$ is a median in triangle $ABC$. So $AQ$ is its isogonal and we are done .

My solution : From the problem Three concurrent radical axes $ \Longrightarrow AQ $ pass through the isogonal conjugate of the complement of $ P $ , so $ MN \parallel BC \Longrightarrow AP $ is A-median of $ \triangle ABC \Longrightarrow AP $ pass through the complement of $ P \Longrightarrow \angle BAQ=\angle CAP $ . Q.E.D

An extension Let $ABC$ be a triangle with $E,F$ are arbitrary points on $CA,AB$. Circle $(ABE),(ACF)$ intersect again at $D$. $M,N$ are midpoints of $BC,EF$. $G$ lies on $BC$ such that $AG\parallel MN$. Prove that $\angle DAB=\angle GAC$.

buratinogigle wrote: An extension Let $ABC$ be a triangle with $E,F$ are arbitrary points on $CA,AB$. Circle $(ABE),(ACF)$ intersect again at $D$. $M,N$ are midpoints of $BC,EF$. $G$ lies on $BC$ such that $AG\parallel MN$. Prove that $\angle DAB=\angle GAC$. My solution : Let $ T \equiv BE \cap CF $ and $ S $ be the reflection of $ T $ in $ M $ . Since $ MN $ passes through the midpoint of $ AT $ ( Newton line theorem ) , so $ A, G, S $ are collinear $ \Longrightarrow AG $ passes through the complement of $ T $ WRT $ \triangle ABC $ , hence from the problem Three concurrent radical axes $ \Longrightarrow AD, AG $ are isogonal conjugate WRT $ \angle BAC $ . Q.E.D

Mr.Buratino wrote: An extension Let $ABC$ be a triangle with $E,F$ are arbitrary points on $CA,AB$. Circle $(ABE),(ACF)$ intersect again at $D$. $M,N$ are midpoints of $BC,EF$. $G$ lies on $BC$ such that $AG\parallel MN$. Prove that $\angle DAB=\angle GAC$. My solution: $MN\cap{CA, AB} = U, V$; $H$ is the midpoint of $CF$ $X, Y$ are the projections of $D$ on $AB, AC$ $K, L$ are the projections of $G$ on $CA, AB$ We have: $\triangle{DBF}$$\sim$$\triangle{DEC}$ $\Rightarrow$ $\frac{DX}{DY} = \frac{BF}{EC}$ (1) On the other hand: $\triangle{HMN}$$\sim$$\triangle{AVU}$ $\Rightarrow$ $\frac{BF}{EC} = \frac{HM}{HN} = \frac{AV}{AU} = \frac{sin\angle{AUV}}{sin\angle{AVU}} = \frac{sin\angle{GAK}}{sin\angle{GAL}} = \frac{GK}{GL}$ (2) (1), (2) $\Rightarrow$ $\frac{DX}{DY} = \frac{GK}{GL}$ and the conclusion follows Q.E.D

Nice solution dear tranquanghuy7198, a little change without sine Let $AZ\perp UV, Z\in UV$ then $\triangle AZU\sim\triangle AKG$ and $\triangle AZV\sim\triangle ALG$ so $\frac{BF}{EC} = \frac{HM}{HN} = \frac{AV}{AU} =\frac{AV}{AZ}.\frac{AZ}{AU} =\frac{AG}{GL}.\frac{GK}{AG}= \frac{GK}{GL}$.

Note that by a standard Yufei/spiral similarity configuration, $\triangle QMB\sim\triangle QCN$. This means that $\angle BMQ=\angle NCQ$, so quadrilateral $AMQC$ is cyclic. Similarly, $ANQB$ is also cyclic. Now perform a $\sqrt{bc}$ inversion $\Phi$ about $A$. Note that $\Phi$ sends $B$ to $C$ and vice versa. Furthermore, $\Phi$ sends $M$ to the point $M'$ on ray $\overrightarrow{AC}$ such that \[AM\cdot AM'=AB\cdot AC\quad\implies\quad AM'=2AC.\]Similarly, this inversion sends $N$ to the point $N'$ such that $AN'=2AB$. As a result, $\Phi$ sends $(ANB)$ to $CN'$ and $(AMC)$ to $BM'$, meaning that the image of $Q$ about $\Phi$ is the centroid of the triangle homothetic to $\triangle ABC$ with scale factor $2$ (i.e. $\triangle AN'M'$). Hence $Q'$ lies on the $A$-median of $\triangle ABC$, meaning that $Q$ lies on the $A$-symmedian, so by the definition of symmedian we have $\angle BAQ=\angle CAP$ as desired. $\blacksquare$

Claim: $ABQN$ and $ACQM$ are cyclic. Proof. By the Spiral Center Lemma, we know $Q$ is the center of the spiral similarity taking $BM$ to $NC$. Thus, $Q$ is also the spiral center between $BN$ and $MC$. But $BM \cap NC = A$, so $A$ lies on $(QBN)$ and $(QMC)$ as desired. $\square$ Now, consider the inversion about $A$ with radius $\sqrt{AM \cdot AC} = \sqrt{AN \cdot AB}$ (equal since $AMN \sim ABC$) followed by a reflection over the $A$-bisector. Notice $B$ swaps with $N$, and $C$ swaps with $M$. Thus, $$P^* = \overline{B^*N^*} \cap \overline{C^*M^*} = (ABN) \cap (ACM) = Q$$(as $P^* \ne A$). The result follows easily. $\blacksquare$ Remark: The inversion I performed was motivated by numerous "potential/desired" spiral similarities, namely $ANP \rightarrow AQB$. This operation is also very similar to $\sqrt{bc}$-inversion. Note: By Ceva's, we know $AP$ is the $A$-median of $ABC$, so $AQ$ is the $A$-symmedian.

Notice that $Q$ is the Miquel point of $BCMN$ so there is a spiral similarity mapping $\overline{MB}$ to $\overline{NC}.$ Hence, the ratio of the distance from $M$ to $\overline{AB}$ and $\overline{AC}$ is $MB/MC=AB/AC.$ Since the ratio of the distance from the midpoint of $\overline{BC}$ to $\overline{AB}$ and $\overline{AC}$ is $AC/AB,$ we see that $\overline{AQ}$ is the $A$-symmedian of $\triangle ABC.$ Since $P$ lies on the $A$-median of $\triangle ABC$ by Ceva, we are done. $\square$

Using Ceva Theorem we have $AP$ is median so we need to prove $AQ$ is symmedian. Let $AQ$ meet circumcircle of $ABC$ at $D$. $\angle NCQ = \angle QPB = \angle QMB$ and $\angle CQN = \angle CPN = \angle MPB = \angle MQB$ so $CNQ$ and $MBQ$ are similar. $\frac{AB}{AC} = \frac{BM}{CN} = \frac{BQ}{QN} = \frac{\sin{BAQ}}{\sin{NAQ}} = \frac{BD}{CD}$ so $AD$ is symmedian. we're Done.

Very nice problem.

[asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.1, xmax = 0.6, ymin = -0.8, ymax = 0.8; /* image dimensions */ pen wwccqq = rgb(0.4,0.8,0); /* draw figures */ draw((-0.5649135165662259,0.536594333380709)--(-0.8852842704686563,-0.32441412262196817), linewidth(0.5)); draw((-0.8852842704686563,-0.32441412262196817)--(0.34335010783265624,-0.3302004697725238), linewidth(0.5)); draw((0.34335010783265624,-0.3302004697725238)--(-0.5649135165662259,0.536594333380709), linewidth(0.5)); draw((-0.752919505397859,0.03132116411857255)--(-0.03190918098033823,0.02792551109933775), linewidth(0.5)); draw((-0.8852842704686563,-0.32441412262196817)--(-0.03190918098033823,0.02792551109933775), linewidth(0.5)); draw((-0.752919505397859,0.03132116411857255)--(0.34335010783265624,-0.3302004697725238), linewidth(0.5)); draw((xmin, -2.938976377952756*xmin-1.1236731473936517)--(xmax, -2.938976377952756*xmax-1.1236731473936517), linewidth(0.5) + wwccqq); /* line */ draw(circle((-0.6122085315419402,-0.22352897588386156), 0.29111539296770506), linewidth(0.5) + blue); draw(circle((-0.052563305306497866,-0.3693858698488498), 0.3978478682645369), linewidth(0.5) + blue); draw((xmin, -7.985780076992614*xmin-3.9746807724376945)--(xmax, -7.985780076992614*xmax-3.9746807724376945), linewidth(0.5) + wwccqq); /* line */ draw(circle((-0.534292834968941,0.03509349622859525), 0.502434787615023), linewidth(0.5) + linetype("4 4") + red); draw(circle((-0.1271159426332545,0.08608123799082845), 0.6281948462534274), linewidth(0.5) + linetype("4 4") + red); /* dots and labels */ dot((-0.5649135165662259,0.536594333380709),linewidth(3pt) + dotstyle); label("$A$", (-0.6821395092458079,0.5800735245046977), NE * labelscalefactor); dot((-0.8852842704686563,-0.32441412262196817),linewidth(3pt) + dotstyle); label("$B$", (-0.9701803985921769,-0.2699294920958601), NE * labelscalefactor); dot((0.34335010783265624,-0.3302004697725238),linewidth(3pt) + dotstyle); label("$C$", (0.37118648806983545,-0.3122884464115025), NE * labelscalefactor); dot((-0.752919505397859,0.03132116411857255),linewidth(3pt) + dotstyle); label("$M$", (-0.8600471173715064,0.05764642127844126), NE * labelscalefactor); dot((-0.03190918098033823,0.02792551109933775),linewidth(3pt) + dotstyle); label("$N$", (-0.03263554307262299,0.06611821214156974), NE * labelscalefactor); dot((-0.3475011702611344,-0.10237541668523897),linewidth(3pt) + dotstyle); label("$P$", (-0.35173966558379643,-0.05813472051764801), NE * labelscalefactor); dot((-0.44030864113663953,-0.4584727983210283),linewidth(3pt) + dotstyle); label("$Q$", (-0.4477532953659194,-0.4167738670567538), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Note that it suffices to prove that $AP$ and $AQ$ are isogonals. Here is an Inversive solution (not gonna lie, but I felt that the Inversive solution was much more motivated than the spiral sim). Firstly, we have, $\measuredangle BQN=\measuredangle BQP+\measuredangle PQN=\measuredangle BMP+\measuredangle PCN=\measuredangle AMC+\measuredangle MCA=\measuredangle MAN=\measuredangle BAN\implies ABQN$ is cyclic and similarly, we also get that $AMQC$ is also cyclic. Now perform an Inversion $\mathbf I(\odot(A,\sqrt{AB\cdot AN}))$ followed by a reflection along the angle-bisector of $\angle BAC$. Firstly, note that $B\leftrightarrow N$. Now using the parallel condition, we get that $\triangle AMN\sim\triangle ABC\implies \dfrac{AM}{AB}=\dfrac{AN}{AC}\implies AC\cdot AM=AB\cdot AN$ from which we get that $M\leftrightarrow C$. Now to finish, note that $BN\leftrightarrow\odot(AMC)$ and $CM\leftrightarrow\odot(ABN)$ and thus $P=BN\cap MC$ swaps with $Q=\odot(BMP)\cap\odot(CNP)$ and we are done. Remark: There actually was no need of spiral similarity at all and just normal dangle chasing suffices, but for the sake of completeness, I am including the (probably?) intended solution in place of the dangle-chasing. Now note that $Q$ is the center of the spiral similarity mapping $\overline{BM}\mapsto \overline{NC}$. So it also maps $\overline{BN}\mapsto \overline{MC}$ and thus as $A=BM\cap NC$, we thus get that the miquel point of $BNCM$, that is $Q$, lies on $\odot(ABN)$ and $\odot(AMC)$ and proceed similarly as the rest above.

First I claim that $Q$ lies on $(ANB)$ and $(AMC)$; this is by angle chasing as $$\measuredangle NQB = \measuredangle PQB + \measuredangle NQP = \measuredangle PMB + \measuredangle NCM = \measuredangle CMB + \measuredangle CNM + \measuredangle NMC = \measuredangle NAB.$$Now the $\sqrt{AM\cdot AC}$ inversion swaps $\overline{BN}$, $(ABN)$ and $\overline{CM}$, $(ACM)$, thus $P, Q$, which implies the result.

By Miquel, $Q$ is the Miquel point of a complete quadrilateral. Hence, $(ANB)$ and $(AMC)$ also pass through $Q$. Thus, we redefine $Q$ as the second intersection of $(ANB)$ and $(AMC).$ Furthermore, by Ceva, $AP$ is the median, so it suffices to show that $Q$ lies on the symmedian. This will set up the barycentric coordinate approach. The purpose of the Miquel step is that $(ANB)$ is a much better circle than say $(BMP)$ since it passes through two vertices. Let $M=(p,1-p,0)$ and $N=(p,0,1-p).$ Consider the intersection of $(ANB)$ with the A-comedian. For $(ANB)$, we have $u=0$ and $v=0$, and plugging in $N=(p,0,1-p)$ gives $$w=b^2p.$$Thus, the equation of $(ANB)$ is $$(x+y+z)(b^2pz)=a^2xy+b^2xz+c^2xy.$$Intersecting this with the A-symmedian, plugging in $(t:b^2:c^2),$ gives $$(t+b^2+c^2)b^2c^2p=a^2b^2c^2+2b^2c^2t$$$$(t+b^2+c^2)p=a^2+2t$$$$t=\frac{a-pb^2-pc^2}{p-2}.$$The important thing about this is that it is symmetric in $b$ and $c$. Thus, if we intersect $(AMC)$ with the A-symmedian, we will get the exact same result. Hence, the second intersection of the two circles lies on the A-symmedian, so we are done.

$Q$ is the miquel point of $ANPK$ so $(ANQB), (AMQC)$ are cyclic. Let $MN\cap (ANQB)=N,K$ and $MN\cap (AMQC)=M,L$ $KB\cap LC=S$ \[\angle CBS=\angle LKS=\angle NKB=\angle NAB=\angle A\]\[\angle SCB=\angle SLK=\angle CLM=\angle CAM=\angle A\]So $SB$ and $SC$ are tangents to $(ABC)$ at $B$ and $C$, respectively. Radical axises of $(KBCL),(KBQA),(CLQA)$ are concurrent which gives that $A,Q,S$ are collinear. $AS$ is symedian so $AQ$ is symedian. $AP$ is median by ceva. This gives that $\angle CAP=\angle QAB $ as desired.$\blacksquare$

We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We let $M = (m, n, 0)$ and $N = (m, 0, n)$ where $m+n=0$. We can see now that $P = (m:n:n)$. We try to find $u$, $v$ and $w$ for the circumcircle $BMP$. We can see from $B$ that $v = 0$. Now we can see from $M$ that $u = c^2n$. Finally we get that $w = \frac{a^2n+b^2m-c^2mn}{n+1}$ from $P$. Now note that our equation for $BMP$ is \[a^2yz + b^2xz + c^2xy = \left((xc^2n + z\frac{a^2n+b^2m-c^2mn}{n+1}\right)(x+y+z)\]Thus symmetrically the equation for $CNP$ is \[a^2yz + b^2xz + c^2xy = \left((xb^2n + y\frac{a^2n+c^2m-b^2mn}{n+1}\right)(x+y+z)\]We can see that the point $Q = \left(\frac{-1}{2n^3(n+1)}, \frac{b^2}{n}, \frac{c^2}{n}\right)$ satisfies both equations. Obviously this point lies on the cevian of the isongonal conjugate of $P$. $\blacksquare$

First observe that $AP$ is the $A$-median from $A$ to $\overline{BC}$. (Find this using Ceva and the fact that $AM/MB = AN/NC$). It therefore suffices to show $Q$ lies on the $A$-symmedian of $\triangle ABC$. We first use a lemma: Lemma: Let $X$ be a point within triangle $ABC$. Then $X$ lies on the $A$-symmedian iff \[\frac{\text{d}(X, AB)}{\text{d}(X, AC)} = \frac{AB}{AC}, \]where $\text{d}(X, AB)$ denotes the distance from point $X$ to line $AB$. Proof: We have that if $A_1$ is where the $A$-symmedian meets line $\overline{BC}$, \begin{align*} \frac{\text{d}(X, AB)}{\text{d}(X, AC)} &= \frac{\text{d}(A_1, AB)}{\text{d}(A_1, AC)} \\ &= \frac{BA_1}{CA_1} \cdot \frac{\sin \angle B}{\sin \angle C} \\ &= \frac{AB^2}{AC^2} \cdot \frac{\sin \angle B}{\sin \angle C} \\ &= \frac{AB}{AC}. \end{align*}The opposite direction is proved similarly. $\square$ Now taking observation of the spiral similarity at $Q$ sending $BM \mapsto NC$, we find that \[\frac{BM}{BQ} = \frac{NC}{NQ} \implies \frac{AB}{AC} = \frac{BM}{NC} = \frac{BQ}{NQ}, \]which by the above lemma is enough to show that $Q$ lies on the $A$-symmedian, as desired. $\blacksquare$

We have that $Q$ is the Miquel Point of $BNCM$, and therefore that $ABQN$ and $ACQM$ are cyclic. Hence the composition of the inversion with centre $A$ and the reflection wrt the angle bisector of $\angle BAC$ that maps $M$ to $C$ (which maps $B$ to $N$, because of the parallel condition) also maps $P$ to $Q$. It follows that $\angle BAQ=\angle CAP$, as desired.

Ceva sine && Ceva && Simmedian && Angle chasing

Performing $\sqrt{bc}$ inversion about $A$ gives that $B$ becomes the midpoint of $AM$, and $C$ becomes the midpoint of $AN$. Then, $Q$ is simply the centroid of $AMN$, so $AQ$ is a median, implying the conclusion.

using ceva yields to the fact that the ray $AY$ stays unchanged so studying the case $D=B$ yields to the fact that $AF$ is the $A$ $symedian$ in $\triangle ABC$ and performing a $\sqrt{AB\cdot AC}$ yields to the result $\blacksquare$

Just use inversion with radius $AM*AC$, which flips around $A$- angle bisector. And notice that $BQNA$ and $AMQC$ are cyclic. Which gives that $P$ and $Q$ are pairs under that inversion.