Just consider points $B'$ and $C'$ which are symmetric to $B$ and $C$ with respect to $A$ and find two pairs of similar triangles $C'BC$, $ABM$ and $BCB'$, $ACN$.

$QP$ and $NM$ are parallel. $\angle{ANM} = \angle{AQC} = \angle{A}$ $\angle{AMN} = \angle{APB} = \angle{A}$ Let $R$ be the intersection of $BM$ and $CN$. Let $X, Y$ be points on $AB, AC$ such that $A$ is the midpoint of $BX$ and $A$ is the midpoint of $YC$ $\triangle{ABC}$ and $\triangle{QAC}$ are similar, then $\triangle{XBC}$ and $\triangle{NAC}$ are similar. Then $\angle{ANC} = \angle{BXC}$ $\triangle{ACB}$ and $\triangle{PAB}$ are similar, then $\triangle{YCB}$ and $\triangle{MAB}$ are similar. Then $\angle{AMB} = \angle{CYB}$ $XCBY$ is a parallelogram. Take $H$ be the midpoint of $BC$. Then $AH$ is parallel to $YB$ and $XC$. Then $\angle{A} = \angle{BXC} + \angle{CYB}$ $\angle{RNM} = \angle{ANM} - \angle{ANC} = \angle{A} - \angle{BXC}$ $\angle{RMN} = \angle{RMN} - \angle{AMB} = \angle{A} - \angle{CYB}$ $\angle{RNM} + \angle{RMN} = 2\angle{A} - \angle{BXC} - \angle{CYB} = \angle{A}$ $\angle{BRC} + \angle{A} = 180$

Let $x=<QCN$ and $y=<PBM$, by similitude we have $\frac{AQ}{AB}=\frac{AC}{BC}=\frac{QC}{AC}$ so $QN=\frac{AB.AC}{BC}$ and $QC=\frac{AC^2}{BC}$ and $<CQN=180-<AQC=180-A$ so in the triangle $QCN$ we have \[\frac{\sin(x-A)}{\sin(x)}=\frac{AC}{AB}\] and by the same way \[\frac{\sin(y-A)}{\sin(y)}=\frac{AB}{AC}\] so \[\sin(x-A)\sin(y-A)=\sin(x)\sin(y)\] so $-\cos(x+y-2A)+\cos(x-y)=2\sin(x)\sin(y)$ and therefore \[\cos(x+y-2A)=\cos(x+y)\] which gives \[x+y=A\] and if $R$ is the intersection point \[<BRC+<BCR=A\]

very easy problem $\widehat{BPM}=\widehat{CQN},\frac{BP}{PN}=\frac{BP}{AN}=\frac{\sin C}{\sin B}=\frac{AQ}{QC}=\frac{QN}{QC}$ so $\widehat{CNQ}=\widehat{CBM}$ let the intersection of $BM$ and $CN$ is $S$. so $BQSN$ is cycle so $\widehat{BAC}=\widehat{BQN}=\widehat{BSN}.$ done!

Barycentric coordinates overkill this problem, yet straightfoward and not messy, took me about 5 minutes. We have similar triangles $PBA$, $QAC$, $ABC$, with area ratios $\frac{c^2}{a^2}, \frac{b^2}{a^2}, 1$ respectively. Therefore we have the barycentric coordinates $P=(0:1-\frac{c^2}{a^2}:\frac{c^2}{a^2}), Q=(0:\frac{b^2}{a^2}:1-\frac{b^2}{a^2}).$ Now note that the area of $MAB$ is the double of $PAB$, and area of $MCA$ is the double of $PCA$. $MBC$ has the same area as $ABC$, but negative, and do the similar for point $N$. One then obtains $M=(-1:2(1-\frac{c^2}{a^2}):\frac{2c^2}{a^2}), N=(-1:\frac{2b^2}{a^2}:2(1-\frac{c^2}{a^2})).$ Lines $BM$ and $CN$ have equations $z=-\frac{2c^2x}{a^2}$ and $y=-\frac{2b^2x}{a^2}$ respectively. The intersection point has un-homogenized coordinates $(1:-\frac{2b^2x}{a^2}:-\frac{2c^2x}{a^2})$, which can be easily verified to satisfy the circumcircle equation $a^2yz+b^2xz+c^2xy=0$. Yay!

Another approach. Let $ O $ be the circumcenter of $ \triangle ABC $. Observe that $ OB \perp AP $, $ OC \perp AQ $. Let $ OB \cap AP = L $, $ OC \cap AQ = K $. Let $ X,Y $ be the midpoints of $ AB,AC $. Let $ XP \cap YQ = S $. Then, we want to show that $ S $ lies on circle $ (AXY) $, i.e, the circle $ \Gamma $ with diameter $ OA $. Note that $ X,K,L,Y $ lie on $ \Gamma $, so by Pascal's Theorem on $ KOLYAX $, we get $ KO \cap YA = C, LO \cap XA = B $ and $ LY \cap KX := D $ are collinear, i.e $ D $ is on $ \overline{BC} = \overline{PQ} $. Considering the hexagon $ KALYSX $, we have the intersections, $ KA \cap SY = Q, LA \cap SX = P $ and $ LY \cap KX = D $ are collinear, so by converse of Pascal's Theorem, $ S $ is on the unique conic passing through $ A,K,L,X,Y $ which is precisely circle $ \Gamma $.

Let $X$ and $Y$ be the midpoints of $AB$ and $AC$, respectively, and let $Z$ be the intersection between $PX$ and $QY$. Also, let $ \gamma = \angle ACB $ and $\beta = \angle ABC$. A quick angle chasing show us that $\angle APQ = \angle AQP = 180 -\alpha -\beta$, and the triangles $ABP$ and $CAQ$ are similar, where we get $\angle YQC = \angle APX = \alpha$ (because both $QY$ and $PX$ are medians of the same corresponding size). Thus, $\angle YPQ=180- \alpha - \beta - \gamma$ and finally: $\angle XZY = \angle QZP = 180 - \angle ZQP - \angle ZPQ = 180 -\alpha - (180 - \alpha - \beta - \gamma)= \beta + \gamma = 180 - \angle A$. So $AXZY$ is cyclic. Now, let $R$ be the intersection of $BM$ and $CN$. A homotety of center $A$ and ratio 2 takes $X$ to $B$, $Y$ to $C$, $Q$ to $N$ and $P$ to $M$. So it takes $Z$ to $R$, which means that $ABRC$ is cyclic as well. This ends the proof.

It's very nice and easy problem.Let $ (BAQ) \cap (CAP) = X $, and let $ AX \cap (ABC)=Y $, $ BM \cap CN=Y $.We have by angles chasing $ AQ=AP $ , $ MY || PX $ and $ NY || QX $ so the triangles $ QXP $ and $ NYM $ are similar hamotetic (with ratio 2) .On the other $ \angle NYM = \angle QXP=180 - \angle BAC $ so we have $ Y $ lies on the circumcircle of the triangle $ ABC $. DONE !

we have $ ABP$ be similar to $ACQ$. So $ BPM$ is similar to $CQN$ and we are done!

If you knew the following problem, then you had a big advantage. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=73134&p=422851#p422851

An interesting note to make: Draw the circumcircle and the tangents to it at $B$ and $C$. Suppose these intersect at $K$. Obviously $\angle ACK = \angle ABC = \angle CAQ$. Thus $AQ || KC$. Similarly, $AP || KB$. Also, $MN || BC$ so there exists a homothety mapping $KBC$ to $AMN$, which implies that $BM$ and $CN$ intersect on the symmedian through $A$.

What an excellent barycentric tutorial problem! Since $PB = c^2/a$ we have $P = (0 : a^2-c^2 : c^2)$, so $M = (-a^2 : {-} : 2c^2)$. Similarly $N = (-a^2 : 2b^2 : {-})$. Thus $\overline{BM} \cap \overline{CN}$ is $(-a^2 : 2b^2 : 2c^2)$ which clearly lies on the circumcircle.

A good and easy IMO problem. I saw the cyclic quadrilateral and then the proof is already done. There are many different ways of approaching the problem. A purely analytical solution might be doable but gory.

Let $BM$ and $CN$ meet at $J.$ From similar triangles $\triangle ABP$ and $\triangle ACQ,$ we have $\dfrac{AP}{BP} = \dfrac{AQ}{QC} \implies \dfrac{PM}{BP} = \dfrac{QM}{QC}.$ Combined with the fact that $\angle BPM = \angle ABC + \angle BCA = \angle QCN,$ we deduce that $\triangle BPM \sim \triangle QCN,$ implying $\angle PBM = \angle QCN.$ Let $\angle PBM = \angle QCN = \theta.$ Again, from $\triangle ABP \sim \triangle ACQ,$ $\dfrac{PM}{BM} = \dfrac{AP}{BP}= \dfrac{AB}{AC}.$ We have that $\angle BPM = \angle ABC + \angle BCA = 180^{\circ} - \angle BAC.$ We then have $\angle BMP = \theta - \angle BAC .$ Similarly, $\angle CNQ =\theta - \angle BAC.$ From sine rule in $\triangle BPM,$ $\dfrac{PM}{\sin \theta} = \dfrac{BP}{\sin (\theta - \angle BAC)}$ and from sine rule in $\triangle CQN,$ $\dfrac{QN}{\sin \theta } = \dfrac{QC}{\sin (\theta - \angle BAC )}.$ Combining them, we see that \[\sin^2(\theta - \angle BAC) = \sin^2 \theta \implies \cos (2(\theta - \angle BAC)) + 1 = \cos^2 \theta \\ \implies 2 \theta = \angle BAC \implies \theta = \dfrac{\angle BAC}{2}.\] Let $BM$ and $CN$ meet at $J.$ From isoceles $\triangle JBC,$ we have that $\angle BJC = 180^{\circ} - 2 \theta = 180^{\circ} - \angle BAC,$ so $(ABJC)$ is cyclic, as desired. $\blacksquare$

ipaper wrote: Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$. My solution : First of all, we notice that $\triangle{CPA} \sim \triangle{CAB}$ and $\triangle{BAQ} \sim \triangle{BCA}$, so $\angle{QPA} = \angle{BAC} = \angle{PQA}$ hence, $\triangle{APQ}$ is isosceles. So, $\angle{BPM} = \angle{CPA} = \angle{BAC}$. This also means $PA = PM = QA = QN$. Clearly, from $PQ \parallel MN$, we get $\angle{AMN} = \angle{APQ} = \angle{BAC}$ Suppose that $X$ is the intersection of $BN$ and $CM$, then we need to prove $\angle{BXM} = \angle{BAC} = \angle{BPM}$ or this means it suffices to prove that $MXPB$ is cyclic and again this implies that we only need to prove that $CP\cdot CB = CX\cdot CM = CA^2$. Now, i will use length chasing to prove $CX\cdot CM = CA^2$. Denote by $a, b, c$ the length of the sides $BC, CA, AB$ respectively. Then, because $\triangle{CPA} \sim \triangle{CAB}$ we have $\frac{AP}{AC} = \frac{BA}{BC}$ so $AP = \frac{bc}{a}$. Now, we compute some length, that is : \[\frac{\frac{MN}{2}}{MA} = \cos \angle{AMN} = \cos \angle{A} = \frac{b^2 + c^2 - a^2}{2bc}\] \[\frac{MN}{2} = \frac{2bc}{a}\cdot\frac{b^2+c^2-a^2}{2bc}\] \[MN = \frac{2b^2+2c^2-2a^2}{a}\] We also have by the law of cosine in $AMC$ : \[CM^2 = AM^2 + AC^2 - 2\cdot AM\cdot AC\cdot\cos{\angle{MAC}}\] \[CM^2 = AM^2 + AC^2 - 2\cdot AM\cdot AC\cdot\cos{\angle{B}}\] But, $\cos{\angle{B}} = \frac{c^2 + a^2 - b^2}{2bc}$. So, \[{CM^2 = \frac{4b^2c^2}{a^2} + b^2 - 2\cdot(\frac{2bc}{a}}\cdot b \cdot{\frac{c^2+a^2-b^2}{2bc}})\] Simplifying this expression \[CM^2 = \frac{2b^2c^2 - a^2b^2 + 2b^4}{a^2}\] Observing that $\triangle{BCX} \sim \triangle{NMX}$, we have $\frac{CX}{CM} = \frac{BC}{BC + MN}$ multiplying both sides with $CM^2$ yields $CX\cdot CM = \frac{BC}{BC + MN}\cdot CM^2$, hence \[CX\cdot CM = \frac{a}{a + \frac{2b^2+2c^2-2a^2}{a}}\cdot \frac{2b^2c^2-a^2b^2-+2b^4}{a^2}\] \[CX\cdot CM = \frac{a^2}{2b^2+2c^2-a^2}\cdot \frac{b^2(2b^2 + 2c^2 - a^2)}{a^2} = b^2 = AC^2\] So, now it is proven, and this problem is solved.

From the alternate segment theorem, the tangent at $B$ is parallel to $AP$. Let $BM$ meet the circumcircle at $X$. Then $B(A,M;P,\infty)=-1\Longrightarrow (A,X;C,B)=-1$ and $AXBC$ is a harmonic quadrilateral. It follows by symmetry that $CN$ also passes through $X$.

I think my solution was not posted yet, and despite using nothing but symmetry, it is quite short Let $X=BM\cap CN$, $Y=AB\cap MN$ and $Z=AC\cap MN$. Then $\triangle MYA$ and $\triangle NAZ$ are both similar to the original triangle $\triangle ABC$. The segments $NC$ and $MB$ are corresponding medians, so comparing the angles gives $\sphericalangle CNA =\sphericalangle BMY$. As $MY$ and $BC$ are parallel, $\sphericalangle BMY = \sphericalangle MBC$. Combining those two equalities, we get \[ \sphericalangle XBQ = \sphericalangle MBC = \sphericalangle BMY = \sphericalangle CNA = \sphericalangle XNQ,\] so the quadrilateral $QBNX$ is cyclic and therefore \[ \sphericalangle BXN = \sphericalangle BQN = \sphericalangle CQA = \sphericalangle BAC, \] where the last equality follows from the similarity of $\triangle ABC$ and $\triangle QAC$. As $\sphericalangle BXN = \sphericalangle BXC$, the above equality implies $BXCA$ is cyclic, which is what we wanted to prove.

What a cute problem.

Very low power solution using the most basic of geometry tools: cyclic quadrilaterals, collinear points, and similar triangles. Denote by T the intersection of circles APC and AQB, X the intersection point of the tangents at B and C to the circucmricle of ABC, K the intersection of AX with the circumcircle of ABC, and R the intersection of BK with AP. 1. Let <BAT = <ACT (given) = x, <CAT = <ABT (given) = y. 2. Angle chase to get BTC = 2A and ATB = ATC = 180-A. 3. Angle chase to get BOC = 2A, BXC = 180-2A, <XBC = <XCB = A so BTOCX is cyclic. 4. <OTK = <OTX = <OBX = 90 (3) 5. <ATB + <BTX = 180-A + BCX = 180 so A, T, X are collinear. 6. A, T, K, X are collinear by definition of K. 7. <APC = <PBA + <BAP = B + x + <TAP = B + x + TCP = B + x + C-x = B + C (angle chase + ATPC is cyclic by definition of P) 8. <ARB = x (angle chase) 9. <APT = <ACT = x (ATPC is cyclic by definition of P) 10. <TAP = <KAR, <APT = <ARK (=ARB) so triangle TAM is similar to triangle KAR. 11. RM/RA = KT/TA 12. KT/TA = 1/2 (4) 13. RM/RA = 1/2 (11+12) 14. PM/PA = 1/2 (by definition of P) 15. P=R (Both P and R lie on the same line, on the same side of P, and satisfy the same length ratios) 16. Line BP = BR passes through K by definition of R 17. Similarly CQ passes through K 18. BP and CQ intersect at K, which by definition lies on the circumcircle of ABC The fact that AT is a symmedian is irrelevant, just that AK passes through X (which implies the symmedian) Beautiful problem IMO (pun intended). I hope this trend of really nice geometry IMO #4's continues.

Denote the intersection of the tangents at $B$ and $C$ as $T$. We first notice $BT \parallel AP$ and $CT \parallel AQ$, as \[\angle CBT = \angle APB = \angle AQC = \angle BCT = \angle A.\] Let $K = AT \cap (ABC)$, implying $(A, K; C, B) = -1$. Since $P$ is the midpoint of $AM$, if we let the infinity point $P_{\infty} = BT \cap AP$, it follows that \[-1 = (A,M;P,P_{\infty}) \overset{B}{=} (A,BM \cap (ABC);C,B),\] so $K = BM \cap (ABC)$, or $K$ lies on $BM$. Analogously, $K$ also lies on $CN$, concluding our proof. $\blacksquare$

let $BM$ and $CN$ intersect at $O$ first, since $BAP=BCA$, $BPA=BAC$, and we can easily prove that $APQ$ is isoceles with $AP=AQ$, and $BAP$, $ACP$, and $BCA$ are similar let $BC=a$, $AB=c$, and $AC=b$ $AP=AQ=\frac{bc}{a}$, $QC=\frac{b^2}{a}$, $BP=\frac{c^2}{a}$, we can easily get that $NQC$ and $BPM$ are similar, and $BCN=BMP$, $MBC=QNC$, $BQN=BON$, $CPM=COM$, and since $AQC=BQN$, $BON=AQC=BAC$, and $BAC+BOC=180$ so $ABOC$ are concyclic

Take a force overlay inversion at $A.$ Our condition inverts to $A,B,BQ\cap CP,C$ is a parallelogram with center $X$ the midpoint of $BC.$ Thus homothety at $A$ gives $X$ lies on the line through $N$ parallel to $AC$ and the line through $M$ parallel to $AB.$ But by Reim's we then have $ABMX,ACNX$ are cyclic which finishes.

Since $\angle{BAP}=\angle{BCA}; \angle{CAQ}=\angle{CBA}$, we have $\angle{AQP}=\angle{APQ}=180-(\angle{B}+\angle{C})$, as well as $\triangle{BAP}\sim \triangle{BCA}; \triangle{CAQ}\sim \triangle{CBA}$ Similarity implies $\frac{BA}{BC}=\frac{BP}{BA}=\frac{AP}{CA}; \frac{CA}{CB}=\frac{CQ}{CA}=\frac{AQ}{BA}$. Generalize to get $\frac{BA}{CA}=\frac{BP}{PM}=\frac{NQ}{CQ}$, satisfying $\triangle{BPM}\sim \triangle{NQC}$, $\angle{CBM}=\angle{CNQ}$. Let $CN\cap BM=K$. We have $B,Q,K,N$ are concyclic as well as $C,P,K,M$. Thus, we get $\angle{BKN}=\angle{CKM}=\angle{BQN}=\angle{CPM}=180-(\angle{B}+\angle{C}); \angle{BKC}=\angle{B}+\angle{C}; \angle{BNC}+\angle{BAC}=180\implies N\in (ABC)$ as desired.

We utilize barycentric coordinates! Set $\triangle ABC$ be the reference triangle. Now, clearly due to the given angle condition, $\triangle ABP \sim \triangle ABC$. Thus, \[\frac{BP}{AB}= \frac{AB}{AC} \implies BP = \frac{c^2}{a}\]from which we have that $PC= \frac{a^2-c^2}{a}$ as well. Thus, $P=(0:a^2-c^2:c^2)$. Similarly, we also have that $Q=(0:b^2:a^2-b^2)$. Since $M$ lies on $AP$ , we have that $M=(t:a^2-c^2:c^2)$. Further, since clearly $[BMC]=[ABC]$, we must have \begin{align*} \frac{t}{t+(a^2-c^2)+c^2} &= -1\\ t &= -t-a^2\\ t &= \frac{-a^2}{2} \end{align*}Thus, we obtain that $M=(-a^2:2(a^2-c^2):2c^2)$. We similarly have that $N=(-a^2:2b^2:2(a^2-b^2))$ as well. Now, let $X=(x:y:z)$. Since $X$ lies on $BM$, \begin{align*} \begin{vmatrix} 0 & 1 & 0\\ -a^2 & 2(a^2-c^2) & 2c^2\\ x & y & z \end{vmatrix}&= 0\\ -a^2z-2c^2x &= 0\\ a^2z=-2c^2x \end{align*}Further, since $X$ also lies on $CN$, \begin{align*} \begin{vmatrix} 0 & 0 & 1\\ -a^2 & 2b^2 & 2(a^2-b^2)\\ x & y & z \end{vmatrix} & =0\\ -a^2y -2b^2x &= 0\\ a^2y &= -2b^2x \end{align*}Thus, $X=(x: \frac{-2b^2}{a^2}x : \frac{-2c^2}{a^2}x) = (a^2:-2b^2:-2c^2)$. Since, \[a^2(-2b^2)(-2b^2)+b^2(a^2)(-2c^2)+c^2(a^2)(-2b^2) = 4a^2b^2c^2 - 2a^2b^2c^2 - 2a^2b^2c^2 =0\]it is quite clear that $X$ lies on the circumcircle of $\triangle ABC$, as desired.

Let $G$ be the intersection of $BM$ and $CN$. Firstly, by the given angle conditions, we get that $\angle AQP = \angle APQ = \angle A$ Also, note that $\triangle ABC \sim \triangle QAC \sim \triangle PBA \implies \frac{QA}{QC} = \frac{PB}{PA}$. Since, $QA = QN$ and $PA = PM$. We have that, $\frac{QA}{QC} = \frac{PB}{PA} \implies \frac{QN}{QC} = \frac{PB}{PM}$ And also $\angle NQC = \angle BPM = 180^{\circ} - \angle A$. So, we get that $\triangle QNC \sim \triangle PBM$ This implies that $\angle QNC = \angle PBM = \angle QBM \implies \square QBNG$ is cyclic. Which means that $\angle BAC = \angle AQP = \angle BQN = \angle BGN = \angle A$. And hence $\square ABGC$ is cyclic.

First, note that the angle conditions tell us that $AP$ is parallel to the tangent to $(ABC)$ at $B$ and that $AQ$ is parallel to the tangent to $(ABC)$ at $C$. Let $\infty_P$ and $\infty_Q$ be the points at infinity along the lines $AP$ and $AQ$, respectively. We have that $(A,M; P,\infty_{P})$ is a harmonic bundle since $P$ is the midpoint of $AM$, and similarly we have $(A,N;Q,\infty_{Q})$ is a harmonic bundle as well. Let $X$ be the intersection of $BM$ with $(ABC)$, distinct from $B$, and let $X'$ be the intersection of $CN$ with $(ABC)$, distinct from $C$. Then we have: \begin{align*} -1&=(A,M;P,\infty_P) \overset{B}{=}(A,X;C,B)=(A,X;B,C) \\ -1&=(A,N;Q,\infty_Q) \overset{C}{=} (A,X';B,C) \end{align*}Thus $X$ and $X'$ must be the same point, so $BM$ and $CN$ meet on the circumcircle, as desired.

storage :p (a) We want to prove that the tangent to $B$ is parallel to $\overline{APM}$, so we need the obtuse angle formed between the tangent and $BC$ to equal $\angle APC$. Because $\angle CAP = \angle A-\angle C$, it's easy to see that $\angle APC = 180-\angle A$. It's trivial that the obtuse angle between the tangent and $BC$ is $180-\angle A$, so we're done with this part. (b) It's well-known that $(AM; P\infty)$ is a harmonic bundle, where $P_\infty$ is the point of infinity on the tangent at $B$. Here, we let $\overline{BM}$ meet the circumcircle at $X$. (c) We have $-1=(AM; P\infty)\stackrel{B}{=}(AX; BC)$, so $ACXB$ is a harmonic quadrilateral. (d) If $CN$ intersects the circumcircle at $Y$, we can get that $(AY; BC)=-1$ too -- therefore, $X$ and $Y$ have to be the same point, implying the conclusion. $\square$

We desire to prove that $BM \cap CN,A , B, C$ cyclic, so it suffices to prove $\angle ABM + \angle ACN = 180^{\circ}$. We write $\angle ABM + \angle ACN = \angle ABC + \angle ACB + \angle CBM + \angle BCN = 180^{\circ} - \angle BAC + \angle CBM + \angle BCN$ so it suffices to prove $\angle CBM + \angle BCN = \angle BAC$. Now we claim $MPB$ is similar to $CQN$. We use $SAS$ similarity. Clearly angles $\angle MPB = \angle CQN = 180^{\circ} - \angle BAC$. We see $CQ : QN = CQ : AQ = AC : AB = AP : PB = MP : PB$, giving the desired similarity. Now we are clearly done since $\angle CQN = \angle PBM$, then $\angle MBP + \angle PBM = \angle BAC$ as desired.

Let $S$ be a point on $(ABC)$ such as $AS$ is $A$ - symmedian of $\triangle ABC;$ $Cx$ be tangent at $C$ of $(ABC)$. Since $\angle{QAC} = \angle{ABC},$ we have $CA$ tangents $(ABQ)$. So $\angle{AQB} = \angle{BAC} = \angle{BCx}$ or $Cx \parallel AQ$. Hence $C(AN, Qx) = - 1 = C(AS, Bx)$ or $C, S, N$ are collinear. Similarly, we have $B, S, M$ are collinear

Denote $K \in (ABC)$ with $(AK;CB) = -1$. Notice \[(AM; PP_{\infty}) \overset{B}{=} (A, BM \cap (ABC); C, B) \implies K \in BM.\] Similarily, $K \in CN$, giving the desired. $\blacksquare$

First, we make the following claim. *** Claim 1. $BM$ and $CN$ intersect at $X$, where $X$ is the unique (by uniqueness of harmonic conjugates) point on $(ABC)$ such that $(AX;BC)=-1$. It now suffices to show that this point $X$ lies on both $BM$ and $CN$. Let $N'=CX\cap AQ$. We make the following claim. *** Claim 2. $Q$ is the midpoint of $AN'$. In other words, $N'=N$. Proof. Let $T$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$. Notice that, \[\angle AQC=180-\angle CAQ-\angle C=180-\angle B-\angle C=\angle A=\angle BCT,\]so $CT\parallel AQ$, which means that we can then get that \[-1=(AX;BC)\overset{B}{=}(AX;TY)\overset{C}{=}(AN';P_{\infty, CT}Q),\]so $(AN';QP_{\infty, CT})=-1$, implying that $Q$ must be the midpoint of $N'$, as desired. Therefore $N=N'$. *** Since $N'\in CX$ and $N=N'$, this means that $X$ lies on $CN$. Similarly, we can prove that $X$ also lies on $BM$, which means that $BM$ and $CN$ both intersect at point $X$, which lies on the circle $(ABC)$. This completes our proof.

Let $B'$ be the reflection of $A$ over $B$ and let $C'$ be the reflection of $A$ over $C$. Let $X = BM \cap CN$. Then $\triangle ABC \sim \triangle MB'A \sim \triangle NAC',$ and $\triangle B'BM \sim \triangle ANC$, so \begin{align*}\measuredangle BXC &= \measuredangle XBC + \measuredangle BCX \\ &= \measuredangle XMN + \measuredangle MNX \\&= \measuredangle CNA + \measuredangle C'NC \\&= \measuredangle C'NA = \measuredangle BAC.\end{align*}Therefore, $X$ lies on $(ABC)$.

Let $J$ be the reflection of $B$ across $A$ and let $K$ be the reflection of $C$ across $A$ so that $JKBC$ is a parallelogram. By the definitions of $P$ and $Q$, we have that $\triangle ABC \sim \triangle PBA \sim \triangle QAC$. So, we also have \[PABM \sim ACBK, \qquad QACN \sim ABCJ.\]Now, letting $X$ be the intersection between lines $BM$ and $CN$, we have that \begin{align*}\angle BXC &= 180^{\circ} - \angle MBP - \angle NCQ \\ &= 180^{\circ} - \angle KBA - \angle JCA = 180^{\circ} - \angle A,\end{align*}as desired.

Reflect $B$ and $C$ about $A$ to get points $X$ and $Y$ respectively. Then $\triangle ANC\sim\triangle BXC$, so that $\angle ACN=\angle BCX$ and thus $\angle XCY=\angle BCN$. Similarly, we also have that $\angle XBY=\angle CBM$. Letting $R$ be the intersection of $BM$ and $CN$, we find that $\angle BRC=180^\circ-\angle BCN-\angle CBM=180^\circ-\angle XCY-\angle CBY=180^\circ-\angle BAC$, so $R$ lies on the circumcircle of $\triangle ABC$ as desired. $\blacksquare$