Let $I_1,I_2$ denote the centers of $\omega_b,\omega_c$ respectively, and let $N',M'$ denote the feet from $I_1,I_2$ to $BC$ respectively. Lemma: $\angle NM'Q = \angle MND$. Proof: $\angle NM'Q = \tfrac12 \widehat{QN} = \angle QNE = \angle MND$. $\blacksquare$ Note that $\angle NQM'=\tfrac12 \widehat{M'N} = \angle DNM'=90-A/2$. Now, by LoS on $\triangle NQM'$, \begin{align*} &\qquad \frac{NQ}{\sin \angle NM'Q} = \frac{M'N}{\sin \angle NQM'} \implies \frac{NQ}{\sin \angle MND} = \frac{M'N}{\sin (90-A/2)}\\ &\implies NQ = \frac{M'N\sin \angle MND}{\sin (90-A/2)}=\frac{M'N}{ND} \cdot \frac{ND\sin \angle MND}{\sin (90-A/2)}. \end{align*}Similarly, \[ MQ = \frac{MN'}{MD} \cdot \frac{MD\sin \angle NMD}{\sin(90-A/2)}. \]But $M'N/ND=MN'/MD$ since $\triangle DMN'\sim \triangle DNM'$ are similar iscoceles triangles, and $ND\sin \angle MND=MD\sin \angle NMD$ by LoS on $\triangle MND$. Therefore, $MP=NQ$.

Let $k$ be the ratio of the similarity between $\triangle DBF \sim \triangle DEC$. Let $r_b$ and $r_c$ denote the inradii of these two triangles. Note that \[ \frac{MP}{NQ} = \frac{2r_b \sin \angle DMP}{2r_c \sin \angle DNQ} = \frac{r_b}{r_c} \div \frac{\sin \angle DMN}{\sin \angle DNM} = \frac{r_b}{r_c} \div \frac{DN}{DM}. \]Obviously $\frac{r_b}{r_c} = k$. But $DN$ and $DM$ correspond to the distance from the vertex $D$ to the incircle, so $\frac{DN}{DM} = k$ as well. Thus $\frac{MP}{NQ} = k \div k = 1$ and we're done.

This was also Canada IMO TST # 2

Let the radii of $\omega_B, \omega_C$ be $r_1, r_2$ respectively. Also let the acute angles made by $(\overline{MN}, \overline{DF}),(\overline{MN},\overline{DE})$ be $\alpha, \beta$. Then it suffices to note by extended law of sines that $$\frac{r_1}{r_2} = \frac{DM}{DN} = \frac{\sin(\beta)}{\sin(\alpha)}$$

Let $F_M$, $F_N$ be the projections of $M$, $N$ on $\overline{BC}$; $I_B$, $I_C$ be the centers of $\omega_B$, $\omega_C$; and suppose $\omega_B$, $\omega_C$ are tangent to $\overline{BC}$ at $X$, $Y$. Note that $\angle CDE = \angle BDF = \angle A$. The desired statement is equivalent to proving $\text{Pow}(M,\omega_C) = \text{Pow}(N,\omega_B)$. Denote $DM = DX = s_B$ and $DN = DY = s_C$. We calculate \begin{align*} \text{Pow}(N,\omega_B) &= NI_B^2 -I_BX^2 \\ &= (NF_N-I_BX)^2 + F_NX^2 - I_BX^2 \\ &= NF_N^2 - 2(NF_N)(I_BX) + F_NX^2 \\ &= NF_N^2 + (F_ND+DX)^2 - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= (NF_N^2 + F_ND^2) + DX^2 + 2 (F_ND)(DX) - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= DN^2 + DX^2 + 2(DN \cos A)(DX) - 2(DN \sin A)(DM \tan \tfrac{A}{2}) \\ &= s_C^2 + s_B^2 + 2s_Bs_C \cos A - 2s_Bs_C \sin A \tan \tfrac{A}{2},\end{align*}which is symmetric in $B$ and $C$ as desired.

Let the incircle of $\triangle BDF$ have radius $r_1$ and hit $BC$ at $D_1$. Similarly assume that incircle of $\triangle DCE$ have radius $r_2$ and hit $BC$ at $D_2$. The key idea is to use the similarity $\triangle DBF \sim \triangle DEC$. Note that this means : $\tfrac {DN}{DM}=\tfrac {r_1}{r_2}=\tfrac {\cos B}{\cos C}$. Let $\angle DMN=\angle FMP= \angle MD_1P=\theta$ and $\angle DNM=\angle ENQ= \angle ND_2Q=\varphi$ Since we have $MP=2r_1\cdot \sin \theta$ and $NQ=2r_2\cdot \sin \varphi$. Hence we aim to prove: $\tfrac {\sin \theta}{\sin \varphi}=\tfrac{\cos C}{\cos B}$. Note that by law of sines in $\triangle DNM$ we have : $$\frac {\sin \theta}{\sin \varphi}=\frac {DN}{DM}=\frac{\cos C}{\cos B}$$ This concludes the proof.$\blacksquare$

Let $b,c$ be the inradii of $\triangle BDF$ and $\triangle CDE$ respectively. By chord length formula, it suffice to show that $$\frac{b}{c}=\frac{\sin\angle QND}{\sin\angle PMD} \iff \frac{b}{c}=\frac{DM}{DN}$$which is true because $\triangle BDF\sim \triangle EDC$

$X = \omega_B \cap BC$ $Y =\omega_C \cap BC$ Note that : $\angle FMP = \angle NMD = \angle MCP = \alpha$ $\angle ENQ = \angle MND = \angle NYQ = \beta$ By Law of Sines in $\triangle MPX, \triangle NQY$ and $\triangle DMN$ : $\frac{MP}{\sin \alpha} = \frac{MX}{\cos A/2}$ $\frac{NQ}{\sin \beta} = \frac{NY}{\cos A/2}$ $\frac{DM}{DN} = \frac{\sin \alpha}{\sin \beta}$ \begin{align*} &\frac{MP}{NQ}\\ &=\frac{MX}{NY}\cdot\frac{\sin \alpha}{\sin \beta}\\ &=\frac{MX}{NY}\cdot\frac{DN}{DM}\\ &=\frac{MX}{NY}\cdot\frac{DY}{DX}\\ &=\frac{\sin A}{\sin A/2}\cdot\frac{\sin A/2}{\sin A}\\ &=\boxed{1} \end{align*}

) Let $\omega_B$ and $\omega_C$ intersect $BC$ at $X,Y$ respectively. Furthermore, define $r_b$ and $r_c$ as the radii of $\omega_B$ and $\omega_C$, respectively. We want to show \[2r_b\sin\angle PXM=2r_c\sin\angle QYN\]From basic angle chasing, we have that \[2r_b\sin\angle PXM=2r_b\sin\angle PMF=2r_b\sin\angle DMN\]and similarly we get $2r_c\sin\angle QYN=2r_c\sin\angle DNM$. Rearranging and applying Law of Sines, it suffices to show that \[\frac{r_b}{r_c}=\frac{DM}{DN} \iff \frac{r_b}{DM}=\frac{r_c}{DN}\]However, this is immediate by the fact that $\triangle BFD\sim \triangle BCA\sim \triangle ECD$, so we are done.

Let $I_B$ and $I_C$ be the centres of $\omega_B$ and $\omega_C$, and let $R$ and $S$ be their respective tangency points with $BC$. Let $X$ be the midpoint of $MN$. It suffices to show that $X$ lies on the radical axis of the two circles. Clearly the midpoint $Y$ of $RS$ lies on the radical axis, so it remains to prove that $XY\perp I_BI_C$. Set $D=(0,0)$ in the Cartesian plane with $BC$ being the $x$-axis. Suppose $R=(-r,0)$ , $S=(s,0)$ and $\angle{RDI_B}=\angle{I_BDM}=\angle{NDI_C}=\angle{I_CDS}=\alpha$. Then, it is easy to compute the following: $$I_B=(-r,r\tan{\alpha}) ,\hspace{5pt}I_C=(s,s\tan{\alpha}) ,\hspace{5pt}M=(-r\cos{2\alpha},r\sin{2\alpha}) ,\hspace{5pt}N=(s\cos{2\alpha},s\sin{2\alpha}) ,\hspace{5pt}X=\left(\frac 12(s-r)\cos{2\alpha},\frac 12(s+r)\sin{2\alpha}\right) ,\hspace{5pt}Y=\left(\frac 12 (s-r),0\right)$$Now we can check the product of the gradients of $XY$ and $I_BI_C$: $$\frac{\frac 12 (s+r)\sin{2\alpha}}{\frac 12(s-r)(\cos{2\alpha}-1)}\times \frac{(s-r)\tan{\alpha}}{s+r}=\frac{\sin{2\alpha}\tan{\alpha}}{\cos{2\alpha}-1}=\frac{2\sin^2{\alpha}}{-2\sin^2{\alpha}}=-1$$Hence, the lines are perpendicular, as required.

Let $I_B$ and $I_C$ be the midpoints of $\omega_B$ and $\omega_C$. Let $\omega_B$ and $\omega_C$ be tangent to $CB$ at $R$ and $S$. Let $X$ and $Y$ be the midpoints of $RS$ and $NM$. Then $X$ lies on the power axis of $\omega_B$ and $\omega_C$. There is a spiral similarity between $FB$ and $CE$ with center $D$. Use complex coordinates with $d=0$ and $i_B=1$. We have $m=\overline{r}, n=ri_C, s=\overline{r}i_C, x=\frac{r+\overline{r}i_C}{2}$ and $y=\frac{\overline{r}+ri_C}{2}$. \[ \frac{y-x}{i_B-i_C}=\frac{r+\overline{r}i_C-\overline{r}-ri_C}{2(1-i_C)}=\frac{(r-\overline{r})(1-i_C)}{2(1-i_C)}=\frac{r-\overline{r}}{2}=i\Im(r) \] We get $XY\perp I_BI_C$. $Y$ lies on the power axis of $\omega_B$ and $\omega_C$.We get: \begin{align*} |YN|\cdot|YQ|=|YM|\cdot|YP|\\ |YN|\cdot(|YN|+|NQ|)=|YM|\cdot(|YM|+|MP|)\\ |YN|+|NQ|=|YM|+|MP|\\ |NQ|=|MP| \end{align*}

Let $BC$ be tangent at $w_B,w_C$ at points $P_B,P_C$ and the second common external tangent at $Q_B,Q_C$. Our first claim is that $Q_BQ_C$ is parallel to $EF$. Let $I_B,I_C$ be the incenters , the triangle $I_BDI_C $ is similar to $BDE$ because $\angle I_BDI_C=\angle BDE=180-\angle A$ and from similarity $\frac{DI_B}{DI_C}=\frac{BD}{DC}$. We eventually we get that $\angle DI_BI_C=90-\angle C$ and $DI_CI_B=90-\angle B$ and that the angle between $BC,I_BI_C$ is $(\angle B -\angle C)/2$ and that of $BC,Q_BQ_C$ is $\angle B-\angle C$ and that $EF,Q_BQ_C$ are parallel. Then $P_C,M,Q_B$ and $P_B,N,Q_C$ are collinear : Again we ge that $MDP_C$ is similar to $BDE$ from where we get that $\angle DMP_C=90-\angle C$ and from $EF//Q_BQ_C$ we have that arc $ MQ_B = 90-\angle C$ thus $P_CM$ passes through $Q_B$. Let $P_CM$ meet $w_C$ at $R_C$ and $P_BN$ ,$w_B$ at $R_B$. We have that $P_CR_C=MQ_B$ because $P_CM\cdot PQ_B=P_CP_B^2=Q_BQ_C^2=Q_BR_C\cdot Q_BP_C$ This shows that the radical axis of $w_B,w_C$ passes through the midpoints of $Q_BP_C$ and $Q_CP_B$ which in turn shows that it passes through the midpoint of $MN$ which is what we were looking for.

Long synthetic solution. Redefine $P,Q$ as the intersection points of $\omega_B,\omega_C$ with $\overline{BC}.$ Let $T=\overline{MQ}\cap\overline{NQ},$ and let $W,X,Y,Z$ denote the midpoints of $\overline{PM},\overline{MN},\overline{NQ},\overline{QP}$ respectively. Additionally, let $I_B$ and $I_C$ denote the incenters of $\omega_B$ and $\omega_C$ respectively. $\textbf{Claim: }$ $PN=QM$ $\emph{Proof: }$ Note that there is spiral similarity centered at $D$ sending $\overline{BF}$ to $\overline{EC},$ so $\triangle DPN\sim\triangle DMQ.$ Therefore, since $DP=DM,$ we have $PN=QM,$ as desired. $\blacksquare$ $\textbf{Claim: }$ $\overline{DT}\perp\overline{I_{B}I_{C}}$ $\emph{Proof: }$ Since $D$ is the Miquel point of quadrilateral $PNMQ,$ the quadrilaterals $DTMP$ and $DTNQ$ are cyclic. But since $\angle DMI_{B}=\angle DPI_{B}=\angle DNI_{C}=\angle DQI_{C}=90^\circ,$ the quadrilaterals $DMI_{B}P$ and $DNI_{C}Q$ are cyclic. Hence, $\angle DTI_{B}=\angle DTI_{C}=90^\circ,$ implying the desired conclusion. $\blacksquare$ $\textbf{Claim: }$ $\overline{TD}$ bisects $\angle PTQ$ $\emph{Proof: }$ Since quadrilateral $DTI_{B}P$ is cyclic, $\angle DTP=\angle DI_{B}P=90^\circ-\frac{1}{2}\angle FDB.$ Similarly, we can show that $\angle DTQ=90^\circ-\angle EDC.$ But it is well-known that $\angle FDB=\angle EDC,$ so we are done. $\blacksquare$ $\textbf{Claim: }$ $\overline{XZ}\perp\overline{I_{B}I_{C}}$ $\emph{Proof: }$ Since $PN=QM,$ the Varignon parallelogram of quadrilateral $PMNQ$ is a rhombus, so $\overline{XZ}\perp\overline{WY}.$ Therefore, since $\overline{DT}\perp\overline{I_{B}I_{C}},$ it suffices to show that $\overline{WY}\parallel\overline{DT}.$ Let $G$ be the intersection point of the perpendicular bisectors of $\overline{PM}$ and $\overline{QN}.$ Since $GP=GM,GN=GQ,PN=MQ,$ we have $\triangle GPN\cong\triangle GMQ,$ so $G$ is the Miquel point of quadrilateral $PNMQ.$ Moreover, $G$ is equidistant from $\overline{PN}$ and $\overline{QM},$ so $G$ lies on $\overline{TD}.$ Note that by properties of spiral similarity, $\angle(\overline{WY},\overline{MQ})=\angle WGM.$ Thus, it suffices to show that $\angle GTQ+\angle WGM=90^\circ.$ Since $\angle GWM=90^\circ,$ we have $$\angle WGM=90^\circ-\angle WMG=90^\circ - \angle PMG=90^\circ-\angle PTG=90^\circ-\angle GTQ,$$as desired. $\blacksquare$ Now remark that $Z$ lies on the radical axis of $\omega_B,\omega_C,$ so $\overline{XZ}$ is the radical axis of $\omega_B,\omega_C.$ This implies that $X$ has equal power with respect to both circles, which is exactly what we wanted to prove.

Let $I_1,I_2$ denote the centers of $\omega_b,\omega_c$ respectively, and let $T,K$ denote the feet from $I_1,I_2$ to $BC$ respectively. I will prove $\text{Pow}(M,\omega_C) = \text{Pow}(N,\omega_B)$. Its easy to see $NT=MK$ and $\angle I_1 TM =\angle ACB , \angle I_2 KM =\angle ABC$ , $\frac{I_1 T}{I_2 K}=\frac{\cos\angle ABC}{\cos\angle ACB}$ We know $NI_1 ^2 - I_1 T^2 =NT^2 - 2NT.I_1 T.\cos\angle I_1 TN = MK^2 - 2MK.I_2 K.\cos\angle I_2 KM = MI_2 ^2 - I_2 K^2 $ We are done

Bashy Solution : Let $r_B$ and $r_C$ denotes inradius of $\omega_B$ and $\omega_C$ respectively. Let $s_B$ and $s_C$ denotes semi-perimeter of $\triangle BFD$ and $\triangle CED$ respectively. We aim to calculate $\tfrac{MP}{NQ}$ and will show that it's indeed $1$. First of all, notice that due to laws of sines on $\triangle O_BMP$ we get $$\frac{r_B}{\sin (\angle O_BMP)}=\frac{MP}{\sin(2\angle O_BMP)}$$which results in $MP=2\cos \angle O_BMP\cdot r_B,$ similarily $NQ=2\cos \angle O_CNQ\cdot r_C$. Finally, using laws of sines once more in $\triangle MDN$ we get $$\frac{MP}{NQ}=\frac{2\cos\angle O_BMP\cdot r_B}{2\cos\angle O_CNQ\cdot r_C}=\left(\frac{\sin\angle NMD}{\sin\angle MND}\right)\left(\frac{r_B}{r_C}\right)=\frac{s_C-a\cos\angle C}{s_B-a\cos\angle B}\cdot\frac{r_B}{r_C}=\tan (\tfrac{\angle FDB}{2}) \cdot \frac{1}{\tan (\tfrac{\angle EDC}{2})}=\tan(\tfrac{\angle A}{2})\cdot\frac{1}{\tan(\tfrac{\angle A}{2})}=1$$as desired. $\qquad \blacksquare$

See my purely synthetic solution on my Youtube channel here: https://www.youtube.com/watch?v=sCnOdrEa1L4

Let $\omega_B, \omega_C$ touch $BC$ at $S,T$ and let $X,Y,K,L$ be midpoints of $MN,ST,SM,NT$. Also let $I_B,I_C$ - incenters of $\omega_B, \omega_C$. $SN=MT$ ( why $\triangle KDN\cong \triangle MDT$ by SAS ), which implies $KXLY$ - rhombus ( why midlines ). Also $KL\parallel I_BI_C$ ( why $ I_BK/KD=I_CL/LD$ ). Hence $XY\perp I_BI_C$ and since $Y$ is on radical axis of $\omega_B, \omega_C$, so is $X$, which concludes by PoP.

IMO 2019 SL G2, parmenides51 wrote: Let $ABC$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $BC, CA$, and $AB$, respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$, respectively. Prove that $MP = NQ$. (Vietnam) Solution. Let $r_B$ and $r_C$ denote inradii of $\omega_B$ and $\omega_C$ respectively. Now, we note that \begin{align} \frac{MP}{NQ} &=\frac{2r_B\cos \angle O_BMP}{2r_C\cos \angle O_CNQ} \\&=\left(\frac{r_B}{r_C}\right)\left(\frac{\cos (90^{\circ}-\angle PMD)}{\cos (\angle QND-90^{\circ})}\right) \\&= \left(\frac{r_B}{r_C}\right)\left(\frac{\cos (90^{\circ}-\angle PMD)}{\cos (90^{\circ}-\angle QND)}\right) \\&=\left(\frac{r_B}{r_C}\right)\left(\frac{\sin \angle PMD}{\sin \angle QND}\right) \\&=\left(\frac{r_B}{r_C}\right)\left(\frac{\sin (180^{\circ}-\angle PMD)}{\sin (180^{\circ}-\angle QND)}\right) \\&=\left(\frac{r_B}{r_C}\right)\left(\frac{\sin \angle DMN}{\sin \angle DNM}\right) \\&= \left(\frac{r_B}{r_C}\right)\left(\frac{DN}{DM}\right) \\&=\left(\frac{r_B}{r_C}\right)\left(\frac{\frac{r_C}{\sin \frac{\angle A}{2}}\sin \left(90^{\circ}-\frac{\angle A}{2}\right)}{\frac{r_B}{\sin \frac{\angle A}{2}}\sin \left(90^{\circ}-\frac{\angle A}{2}\right)}\right) \\&= \left(\frac{r_B}{r_C}\right)\left(\frac{r_C}{r_B}\right) \\&=1 \end{align}where $(6)\rightarrow(7)$ follows from Sine rule applied to $\triangle DMN$, $(8)$ follows from Orthic triangle lemma, and facts like $AI=\tfrac{r}{\sin \frac{\angle A}{2}}$, etc. $\quad\blacksquare$

unfortunately missed the genius solution in #20 Use homothety; we can calculate $\frac{DM}{DN}$ and by LoS and tangency this gives $MP=NQ$. Done!

hahahahahahahahahahahahahahahahahahhahahahahahahahhahahahahahaha (grant wrote this on my account) We begin with the following lemma Lemma. Let $ABC$ be a triangle with $AB=AC$. Let $D$ be on $BC$ and $E,F$ be on $AC,AB$ with $CD=DE$ and $BD=DF$.Then, the midpoints of $AD$, $EF$, and $BC$ are collinear. Proof. Animate $D$ on $BC$, it is clear that $E,F$ have degree $1$ so the midpoint of $EF$ has degree $1$. Similarly the midpoint of $AD$ has degree $1$ and the midpoint of $BC$ has degree $0$. Thus we take $D\equiv B,C$, and the midpoint of $BC$ to finish by mmp. $\square$ Now, going back to the original problem, it suffices to show that the midpoint $K$ of $MN$ is on the radical axis. Also, the midpoint $R$ nof $XY$ is on the radical axis where $X,Y$ are tangency points with $BC$. By the lemma, if $MX\cap NY=H$ and $L$ is the midpoint of $DH$, then $L,K,R$ are collinear, so it suffices to show that $L$ is on the radical axis of $\omega_B$ and $\omega_C$. This, however, is true by radical axis on $\omega_B$, $\omega_C$, and the point circle at $D$. $\blacksquare$

ihatemath123 wrote: Proof. Animate $D$ on $BC$, it is clear that $E,F$ have degree $1$ so the midpoint of $EF$ has degree $1$. Similarly the midpoint of $AD$ has degree $1$ and the midpoint of $BC$ has degree $0$. Thus we take $D\equiv B,C$, and the midpoint of $BC$ to finish by mmp. $\square$ The midpoint of two degree $1$ points is not guaranteed to have degree $1$.

Yep I (I mean Benny) am bad. Here’s a fixed proof of the claim. Let $I$ be midpoint of $BC$, $H$ be midpoint of $EF$, and $G$ midpoint of $AD$. Furthermore, let $J=BE\cap CF$ and $K$ be the midpoint of $AJ$. By Newton-Gauss, $I$, $H$, $K$ are collinear, hence it sufficient to show $JD\parallel GI$ by homothety. This is true since \[\measuredangle DIG= \measuredangle GDI = \measuredangle ADC= \measuredangle AFC= \measuredangle BFJ= \measuredangle BDJ\] Also this makes it elem and I use point circle

WLOG $AC>AB$ (this doesn't really matter but it makes the explanation more smooth). The idea here is that $\omega_c$ is "larger" than $\omega_b$, but the arc $NQ$ has a smaller measure than arc $MP$ to compensate. Thus we can essentially split the problem into two parts. First, we find the ratio between the radiuses of $\omega_b$ and $\omega_c$. Then, we find the ratio of the sines of half the arc measures, and show that these multiply to 1. This would solve the problem. \begin{claim} We have $$\frac{R_{\omega_c}}{R_{\omega_b}}=\frac{\cos\gamma}{\cos\beta}.$$\end{claim} Note that $ABDE$ is cyclic, so there is a homothety at $C$ followed by a reflection across the angle bisector that sends $\triangle ACB$ to $\triangle DCE$. Since $$\frac{CD}{AC}=\cos\gamma,$$this homothety has ratio $\cos\gamma$, so $R_{\omega_c}=r\cos\gamma$. Similarly, $R_{\omega_b}=r\cos\beta$, which shows the claim. Now, we find the "ratio of sines of half the arc measures" we were looking for. Since $NE$ is tangent to $\omega_c$, this ratio is equal to $$\frac{\sin\angle ENQ}{\sin\angle FMP}=\frac{\sin\angle DNM}{\sin\angle DMN}=\frac{DM}{DN}.$$However, $$DM=\frac{DB+DF-FB}{2}=\frac{c\cos\beta+b\cos\beta-a\cos\beta}{2}=\cos\beta\cdot \frac{b+c-a}{2}.$$Similarly, $$DN=\cos\gamma\cdot \frac{b+c-a}{2}.$$Thus, $$\frac{DM}{DN}=\frac{\cos\beta}{\cos\gamma},$$which is the reciprocal of the ratio of radii, so we are done.

Unlike post #53 this is actually me now Because $\angle FDB = \angle EDC$, it follows that the ratio of the radius of $\omega_B$ to the radius of $\omega_C$ is $\frac{MD}{DN}$. Now, letting $X$ and $Y$ be the tangency points of $\omega_B$ and $\omega_C$ with $\overline{BC}$, it follows that \[ \frac{MP}{NQ} = \frac{\text{radius} (\omega_B) \cdot \sin ( \angle MXP )}{\text{radius} (\omega_C) \cdot \sin ( \angle NYQ )} = \frac{MD}{DN} \cdot \frac{\sin ( \angle DMN )}{\sin (\angle DNM)} = 1.\]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.465219736921475, xmax = 2.4572223583383397, ymin = -4.131866009456642, ymax = 1.1208129987426592; /* image dimensions */ pen qqwwzz = rgb(0,0.4,0.6); pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw((-1.436498872725946,0.40140832631925505)--(-2.8555447851977687,-2.2932873258036057), linewidth(0.7) + qqwwzz); draw((-2.8555447851977687,-2.2932873258036057)--(1.2688308139529472,-2.327232803973365), linewidth(0.7) + qqwwzz); draw((1.2688308139529472,-2.327232803973365)--(-1.436498872725946,0.40140832631925505), linewidth(0.7) + qqwwzz); draw((-1.436498872725946,0.40140832631925505)--(-1.4587720556310186,-2.3047833976476415), linewidth(0.7) + blue); draw((-2.8555447851977687,-2.2932873258036057)--(-0.7926360753077117,-0.2480025458385933), linewidth(0.7) + blue); draw((1.2688308139529472,-2.327232803973365)--(-1.9741056144780595,-0.6194793960715617), linewidth(0.7) + blue); draw((-1.9741056144780595,-0.6194793960715617)--(-1.4587720556310186,-2.3047833976476415), linewidth(0.7) + ccqqqq); draw((-1.4587720556310186,-2.3047833976476415)--(-0.7926360753077117,-0.2480025458385933), linewidth(0.7) + ccqqqq); draw(circle((-2.0886459647031916,-1.8347001573988362), 0.4648833474369896), linewidth(0.7) + ccqqqq); draw(circle((-0.4720251699614325,-1.593345575866262), 0.7195348249648241), linewidth(0.7) + ccqqqq); draw((-1.6440820352406826,-1.6987610557942068)--(-1.156553700140246,-1.3716452021444483), linewidth(0.7) + blue); draw(circle((-0.788013908035324,-1.6610761378174324), 2.1620302759755687), linewidth(0.7) + qqwwzz); draw((-2.0886459647031916,-1.8347001573988362)--(-1.6440820352406826,-1.6987610557942068), linewidth(0.7) + blue); draw((-2.0886459647031916,-1.8347001573988362)--(-2.3829807767200273,-2.194538378213741), linewidth(0.7) + blue); draw((-0.4720251699614325,-1.593345575866262)--(-1.156553700140246,-1.3716452021444483), linewidth(0.7) + blue); draw((-0.4720251699614325,-1.593345575866262)--(-0.4176549586609028,-0.8758678797249145), linewidth(0.7) + blue); draw((-0.4720251699614325,-1.593345575866262)--(-0.4779470667117702,-2.3128560312983137), linewidth(0.7) + blue); draw((-2.0886459647031916,-1.8347001573988362)--(-2.0924720355047097,-2.2995677599551367), linewidth(0.7) + blue); draw((-2.3829807767200273,-2.194538378213741)--(-1.6440820352406826,-1.6987610557942068), linewidth(0.7) + ccqqqq); draw((-0.4176549586609028,-0.8758678797249145)--(-1.156553700140246,-1.3716452021444483), linewidth(0.7) + ccqqqq); draw((-1.9741056144780595,-0.6194793960715617)--(-0.7926360753077117,-0.2480025458385933), linewidth(0.7) + ccqqqq); /* dots and labels */ dot((-1.436498872725946,0.40140832631925505),dotstyle); label("$A$", (-1.4931395418506677,0.48825897414250724), NE * labelscalefactor); dot((-2.8555447851977687,-2.2932873258036057),dotstyle); label("$B$", (-3.1504944029878994,-2.407846216918973), NE * labelscalefactor); dot((1.2688308139529472,-2.327232803973365),dotstyle); label("$C$", (1.2975399794759466,-2.4388537671444706), NE * labelscalefactor); dot((-1.4471850279001193,-0.8969595278228518),linewidth(4pt) + dotstyle); label("$H$", (-1.7413561523719848,-0.7706475650126973), NE * labelscalefactor); dot((-1.4587720556310186,-2.3047833976476415),linewidth(4pt) + dotstyle); label("$D$", (-1.4807365217558826,-2.6450552771895704), NE * labelscalefactor); dot((-0.7926360753077117,-0.2480025458385933),linewidth(4pt) + dotstyle); label("$E$", (-0.7675628663057479,-0.20010864086354052), NE * labelscalefactor); dot((-1.9741056144780595,-0.6194793960715617),linewidth(4pt) + dotstyle); label("$F$", (-2.1194920566373074,-0.5721992435695124), NE * labelscalefactor); dot((-2.0886459647031916,-1.8347001573988362),linewidth(4pt) + dotstyle); label("$X$", (-2.1070890365425226,-1.750486152138423), NE * labelscalefactor); dot((-0.4720251699614325,-1.593345575866262),linewidth(4pt) + dotstyle); label("$Y$", (-0.4264798136991617,-1.7532770918678257), NE * labelscalefactor); dot((-1.6440820352406826,-1.6987610557942068),linewidth(4pt) + dotstyle); label("$M$", (-1.7357742729406946,-1.4954484010109347), NE * labelscalefactor); dot((-1.156553700140246,-1.3716452021444483),linewidth(4pt) + dotstyle); label("$N$", (-1.1334519591019039,-1.6458363206501384), NE * labelscalefactor); dot((-2.3829807767200273,-2.194538378213741),linewidth(4pt) + dotstyle); label("$P$", (-2.528791719765211,-2.1466069557463855), NE * labelscalefactor); dot((-0.4176549586609028,-0.8758678797249145),linewidth(4pt) + dotstyle); label("$Q$", (-0.682293404125694,-0.8416551152381949), NE * labelscalefactor); dot((-0.4779470667117702,-2.3128560312983137),linewidth(4pt) + dotstyle); label("$T_C$", (-0.5070994443152639,-2.663659807324869), NE * labelscalefactor); dot((-2.0924720355047097,-2.2995677599551367),linewidth(4pt) + dotstyle); label("$T_B$", (-2.1194920566373074,-2.657458297279769), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the center of $\omega_B$ be $X$ and the center of $\omega_C$ be $Y$. Also let $T_B$ and $T_C$ be the tangency points of $\omega_B$ and $\omega_C$ to $\overline{BC}$. Note that we have, \begin{align*} \frac{MP}{NQ} &= \frac{2r_1\cos(\angle XMP)}{2r_2\cos(\angle YNQ)}\\ &= \frac{r_1\cos(90 - \angle PMD)}{r_2 \cos(90 - \angle QNE)}\\ &= \frac{r_1 \sin(\angle PMD)}{r_2 \sin(\angle QNE)}\\ &= \frac{r_1 \sin(180 - \angle NMD)}{r_2 \sin(180 - \angle MND)}\\ &= \frac{r_1 \sin(\angle NMD)}{r_2 \sin (\angle MND)}\\ &= \frac{r_1 \cdot ND}{r_2 \cdot MD}\\ &= \frac{r_1 \cdot r_2 \cdot \tan(\angle NYD)}{r_1 \cdot r_2 \cdot \tan(\angle MXD)}\\ &= \frac{\tan(\angle DYT_C)}{\tan(\angle DXT_B)}\\ &= \frac{\tan(\angle XDT_B)}{\tan(\angle YDT_C)}\\ &= \frac{\tan\left(\frac{1}{2} \angle FDB \right)}{\tan\left( \frac{1}{2} \angle CDE \right)}\\ \end{align*}However note that, \begin{align*} \angle FDB &= 180 - \angle FDC\\ &= \angle A\\ &= 180 - \angle EDB\\ &= \angle EDC \end{align*}and hence the ratio is simply $1$, as desired. $\square$

Let $r_b$, $r_c$ be the radii of $\omega_b$, $\omega_c$. Thus we have \[\frac{MP}{NQ} = \frac{r_b \cdot \sin (180-\angle DMN)}{r_c \cdot \sin (180-\angle DNM)} = \frac{r_1/r_2}{DM/DN} = 1\] from Law of Sines and the similarity $\triangle DFB \sim \triangle DCE$. $\blacksquare$

Let $r_{B,C}$ and $O_{B,C}$ be the radii and center, respectively, of $\omega_{B,C}$, respectively. Note that, by LoS: \[MP=2r_B\cos{\angle PMO_B}=2r_B\sin{\angle NMD}\]\[NQ=2r_C\cos{\angle QNO_C}=2r_C\sin{\angle MND}\] Note that: \[r_B=MD\tan{\angle BDF/2}\]\[r_C=DN\tan{\angle CDE/2}.\]By angle chasing on cyclic quads $BDHF$, $CDHE$ ($H$ is orthocenter): \[\angle BDF=\angle BHF=\angle CHE=\angle CDE.\]Therefore, using LoS: \[r_B/r_C=MD/DN=\sin{\angle MND}/\sin{\angle NMD}.\] Therefore: \[MP/NQ=1\]$\square$

Let the incenters of $\Delta{BFD}$ and $\Delta{CED}$ be $I_1$ and $I_2$ and their inradii be $r_1$ and $r_2$ respectively. Then $\frac{MP}{NQ} = \frac{2 r_1 \sin{\widehat{MP}}}{2 r_2 \sin{\widehat{NQ}}} \overset{\text{(Alt Segt Thm)}}{=} \frac{r_1 \sin{\angle{DMN}}}{r_2 \sin{\angle{DNM}}} \overset{\text{(LOS)}}{=} \frac{r_1 DN}{r_2 DM} \overset{\Delta{DI_1M} \overset{\text{(AA)}}{\sim} \Delta{DI_2N} }{=} 1$ as desired.

trig? more like complex numbers.

Let $I_1$ be the incenter in $\triangle BFD$ and $I_2$ be the incenter in $\triangle DEC$. Let $I_1M = I_1P = r_b$ and $I_2N = I_2Q = r_c$. Also let $\angle PI_1M = 2x$ and $\angle NI_2Q = 2y$, where obviously $\angle I_1MP = \angle I_1PM = 90 - x$ and $\angle I_2NQ = \angle I_2QN = 90 - y$. Also by some angle chase we get that $\angle PMD = x$, $\angle DMN = 180 - x$, $\angle DNM = y$. Now by law of sines on $\triangle MPI_1$ we get that $MP = \frac{\sin \angle PI_1M \cdot I_1M}{\sin \angle I_1MP} = \frac{\sin 2x \cdot r_b}{\sin (90 - x)} = \frac{2\sin x \cdot \cos x \cdot r_b}{\cos x} = 2\sin x \cdot r_b = 2\sin (180 - x) \cdot r_b = 2\sin \angle DMN \cdot r_b$. Now by law of sines on $\triangle NQI_2$ we get that $NQ = \frac{\sin \angle NI_2Q \cdot NI_2}{\sin \angle NQI_2} = \frac{\sin 2y \cdot r_c}{\sin (90 - y)} = \frac{2\sin y \cdot cos y \cdot r_c}{\cos y} = 2\sin y \cdot r_c = 2\sin \angle DNM \cdot r_c$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{2\sin \angle DMN \cdot r_b}{2\sin \angle DNM \cdot r_c} = \frac{\sin \angle DMN \cdot r_b}{\sin \angle DNM \cdot r_c}$. Now by law of sines on $\triangle DNM$ we get that $\frac{\sin \angle DMN}{\sin \angle DNM} = \frac{DN}{DM}$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{DN}{DM} \cdot \frac{r_b}{r_c}$. Also $\triangle DBE \sim \triangle DEC$ $\Rightarrow$ $\triangle DI_1M \sim \triangle DI_2N$ $\Rightarrow$ $\frac{DN}{DM} = \frac{r_c}{r_b}$ $\Rightarrow$ $\frac{MP}{NQ} = \frac{DN}{DM} \cdot \frac{r_b}{r_c} = \frac{r_c}{r_b} \cdot \frac{r_b}{r_c} = 1$ $\Rightarrow$ MP = NQ and we are ready.

$ $

Okayy let’s bashh! We shall show that $Pow_{\omega_{C}}(M)=Pow_{\omega_{B}}(N)$ (which is equivalent to proving the problem statement). Let $r$ be the inradius of $\triangle ABC$. Let $O_{1}$ be the incentre of $\triangle CED$. Let the foot of perpendicular from $O_{1},M$ on $BC$ be $X,Y$ respectively. Observe that $\triangle DBF \sim \triangle ABC \sim \triangle DEC$. Therefore, (skipping a few details) we get $DM=(s-a)cosC=DX, DN=(s-a)cosB,DY=DMcosA=cosAcosB(s-a) $, $MY=MDsinA=sinAcosB(s-a),O_{1}X=rcosC$ Hence, $$Pow_{\omega_{C}}(M)=MO_{1}^2-(rcosC)^2 $$$$=(s-a)^2(cosAcosB+cosC)^2 + (rcosC-sinAcosB(s-a))^2-r^2cos^2C$$$$=(s-a)^2cos^2Acos^2B+(s-a)^2cos^2C+2cosAcosBcosC(s-a)+(s-a)^2sin^2Acos^2B-2r(s-a)sinAcosBcosC$$$$=(s-a)^2(cos^2Acos^2B+cos^2C+sin^2Acos^2B)+2(s-a)cosBcosC(cosA-rsinA)$$$$=(s-a)^2(cos^2B+cos^2C)+ 2(s-a)cosBcosC(cosA-rsinA)$$which is symmetric in $B$ and $C$, so we are done.$\blacksquare$

Very nice and simple Claim: $\color{blue}{MP=NQ}$ Proof: Note that we have, $BF=a \cos B,FD=b \cos B,DB=c \cos B$ and $\angle{FDB}=\angle {A}$. Also, $EC=a \cos C,CD=b \cos C, DE=c \cos C$ and $\angle{EDC}=\angle {A}$. Clearly, $\Delta BFD \sim \Delta ECD$. Denote the inradii of $\Delta BFD$ and $\Delta ECD$ by $r_{1}$ and $r_{2}$ respectively. Thus, $\boxed{\frac{r_1}{r_2}=\frac{\cos B}{\cos C}}$. Let the incircles of triangles $BDF$ and $CDE$ be tangent to $BC$ at $X$ and $Y$ respectively. By alternate segment theorem, $\angle PXM=\alpha=\angle{PMF}=\angle{DMN}$ and $\angle{QYN}=\beta=\angle{QNE}=\angle{DNM}$. Denote the semi-perimeter of $\Delta ABC$ by $s$. Now note that, $DM=(s-a)\cos B$ and $DN=(s-a)\cos C$ implying, $\boxed{\frac{DM}{DN}=\frac{\cos B}{\cos C}}$. Thus by sine rule in triangle $DMN$ we have, $$\frac{DM}{DN}=\frac{\sin \angle DNM}{\sin \angle DMN}=\frac{\sin \beta}{\sin \alpha} \implies \boxed{\frac{\sin \beta}{\sin \alpha}=\frac{\cos B}{\cos C}}$$Now note that $MP=2r_{1}\sin \alpha$ and $NQ=2r_{2}\sin \beta$ implying, $\frac{NQ}{MP}=\left(\frac{r_{2}}{r_{1}}\right)\left(\frac{\sin \beta}{\sin \alpha}\right)=\frac{\cos C.\cos B}{\cos B.\cos C}=1 \implies \boxed{MP=NQ}$ as desired. $\blacksquare$ ($\mathcal{QED}$)