FIXED:$ABC$ be an acute triangle.The lines $l1$ and $l2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively.The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l1$ and $l2$ at the points $E$ and $F$, respectively.If $D$ is the intersection point of the lines $EF$ and $MC$ prove that $\angle ADB = \angle EMF$.

Hello! Suppose that $AG\perp BC$ και $BH\perp CA$ with $G\in l_{2}$ and $H\in l_{1}$. We will first show that $CM\perp EF$.We will use vectors.We know that $\vec{CM}=\frac{\vec{CA}+\vec{CB}}{2}$. Also $\vec{EF}=\vec{MF}-\vec{ME}$.Consequently $2\vec{CM}\cdot \vec{EF}=(\vec{CA}+\vec{CB})(\vec{MF}-\vec{ME})=\vec{CA}\cdot \vec{MF}-\vec{CA}\cdot \vec{ME}+\vec{CB}\cdot \vec{MF}-\vec{CB}\cdot \vec{ME}$. Also $\vec{CA}\cdot \vec{ME}=\vec{CB}\cdot \vec{MF}=0$. Thus $\displaystyle{2\vec{CM}\cdot \vec{EF}=\vec{CA}\cdot \vec{MF}-\vec{CB}\cdot \vec{ME}}$. Since $\displaystyle{\vec{MF}=\frac{1}{2}\vec{AG}}$ we have $\displaystyle{\vec{CA}\cdot \vec{MF}=\frac{1}{2}\vec{CA}\cdot \vec{AG}=\frac{(CA)(AG)\cos (90+\hat{C})}{2}\overset{(AG)=\frac{(AB)}{\sin \hat{B}}}=-\frac{(CA)(AB)\sin \hat{C}}{2\sin \hat{B}}=-\frac{(AB)^{2}}{2}}$. (where we used the sine law in the last equality). Also $\displaystyle{\vec{CB}\cdot \vec{ME}=\frac{1}{2}\vec{CB}\cdot \vec{BH}=\frac{(CB)(BH)\cos (90+\hat{C})}{2}\overset{BH=\frac{(AB)}{\sin \hat{A}}}=-\frac{(CB)(AB)\sin \hat{C}}{2\sin \hat{A}}=-\frac{(AB)^2}{2}}$ where we used the sine law again. We conclude that $\displaystyle{2\vec{CM}\cdot \vec{EF}=-\frac{(AB)^2}{2}+\frac{(AB)^2}{2}=0\Leftrightarrow \vec{CM}\perp \vec{EF}}$ and we are done. $\rule{430pt}{1pt}$ Back to the initial problem,we have $\angle{MDF}=\angle{MBF}$ thus $MDFB$ is cyclic. Also $\angle{MDE}=\angle{MAE}$ thus $MAED$ is also cyclic. Hence $\angle{MAD}=\angle{MEF}$ and $\angle{MFE}=\angle{DBA}$ thus $\triangle{ADB}\simeq \triangle{EMF}$ and we are done.

huynguyen wrote: FIXED:$ABC$ be an acute triangle.The lines $l1$ and $l2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively.The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l1$ and $l2$ at the points $E$ and $F$, respectively.If $D$ is the intersection point of the lines $EF$ and $MC$ prove that $\angle ADB = \angle EMF$. Since $CB\perp MF$ and $CA\perp EM$ we have $MB^2+CF^2=FB^2+CM^2(1)$ and $MA^2+CE^2=EA^2+CM^2(2)$.Substracting $(2)$ from $(1)$ we obtain $CF^2-CE^2=FB^2-EA^2=(FB^2+BM^2)-(EA^2+AM^2)=MF^2-ME^2$,which implies $CE^2+MF^2=CF^2+ME^2$,which is equivalent to $CM\perp EF$. From here it follows that quadrilaterals $EDMA$ and $DFBM$ are inscriptible.So $\widehat{ADM}\equiv\widehat{AEM}\equiv\widehat{CAB}$ and $\widehat{MDB}\equiv\widehat{MFB}\equiv\widehat{CBA}$,implying that $m(\widehat{ADM})=180^0-m(\widehat{ACB})=m(\widehat{EMF})$. I think this problem is too easy for a JBMO.

huynguyen wrote: FIXED:$ABC$ be an acute triangle.The lines $l1$ and $l2$ are perpendicular to $AB$ at the points $A$ and $B$, respectively.The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l1$ and $l2$ at the points $E$ and $F$, respectively.If $D$ is the intersection point of the lines $EF$ and $MC$ prove that $\angle ADB = \angle EMF$. My solution : From A beautiful perpendicularity / MB-232 $ \Longrightarrow CM \perp EF $ , so we get $ A, M, D, E $ are concyclic and $ B, M, D, F $ are concyclic , hence $ \angle ADB=\angle ADM+\angle BDM=\angle AEM+\angle BFM=\angle EMF $ . Q.E.D

My solution: Let $ME\cap AC\equiv W, MF\cap BC\equiv S$ Since $M,W,C,S$ cyclic $\Rightarrow \angle BCA=\angle MCA+\angle MCB=\angle BMF+\angle EMA=\angle MFE+\angle MEF$ $(1)$ By some chasing, easy to see $\frac{sin\angle MCA}{sin\angle MCB}=\frac{MW}{MS}=\frac{MF}{ME}=\frac{sin\angle MEF}{sin\angle MFE}$ $(2)$ From $(1)$ and $(2)$, combine some trigonometric-chasing $\Rightarrow \angle MCA=\angle MEF \Rightarrow (C,D,W,E)$ cyclic $\Leftrightarrow MC\perp EF$. So, $\angle BDA=\angle BDM+\angle MDA=\angle EMF$ Done!

Let $X$ be the midpoint of $EF$. Since $M$ is the midpoint of $AB$, we have $MX\parallel l_1\parallel l_2$ , so $MX\perp AB$. Now, using $\vec{a} \cdot \vec{b} = 0 \iff a\perp b$ multiple times, we have: $\vec{CM} \cdot \vec{EF} = $ $= \frac{1}{2} (\vec{CA}+\vec{CB}) \cdot (\vec{EM}+\vec{MF}) =$ $ = \frac{1}{2} (\vec{CA} \cdot \vec{EM}+\vec{CA} \cdot \vec{MF} + \vec{CB} \cdot \vec{EM} + \vec{CB} \cdot \vec{MF}) =$ $ = \frac{1}{2} (0 +\vec{CA} \cdot \vec{MF} + \vec{CB} \cdot \vec{EM} + 0) =$ $ = \frac{1}{2} ( (\vec{CB} + \vec{BA}) \cdot \vec{MF} + (\vec{CA} + \vec{AB}) \cdot \vec{EM}) =$ $ = \frac{1}{2} ( \vec{CB} \cdot \vec{MF} + \vec{BA} \cdot \vec{MF} + \vec{CA} \cdot \vec{EM} + \vec{AB} \cdot \vec{EM}) =$ $ = \frac{1}{2} ( 0 + \vec{BA} \cdot \vec{MF} + 0 + \vec{AB} \cdot \vec{EM}) =$ $ = \frac{1}{2} \vec{BA} \cdot ( \vec{MF} - \vec{EM}) =$ $ = \vec{BA} \cdot \frac{1}{2} ( \vec{MF} + \vec{ME}) =$ $ = \vec{BA} \cdot \vec{MX} =$ $=0$ so $CM \perp EF$ and the rest is the same as in the previous proofs.

[asy][asy] import graph; size(15cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-1.9153608146613847,xmax=1.0425091723606323,ymin=0.7844814188011924,ymax=3.196675390976039; pen qqwuqq=rgb(0.,0.39215686274509803,0.); pair A=(-1.1029451544093976,2.9498075861150994), B=(-1.38,0.98), C=(0.39875345723651445,2.1892068503249074), M=(-1.2414725772046986,1.9649037930575497), F=(-0.4864968506556744,0.8543281431032907), D=(-0.6494410588969103,2.0458648781559896), G=(-0.8273354283999673,1.3557045605122202), H=(-0.8161884583812093,2.8045671548459046); D((-1.118012803061264,2.8426794521310383)--(-1.010884669077203,2.827611803479172)--(-0.9958170204253366,2.934739937463233)--A--cycle,qqwuqq); D((-0.8650698856738536,2.7080576879909013)--(-0.7685604188188505,2.659176260698257)--(-0.7196789915262063,2.7556857275532605)--H--cycle,qqwuqq); D((-0.7378683042516357,1.416524827064004)--(-0.7986885708034195,1.5059919512123356)--(-0.8881556949517511,1.4451716846605518)--G--cycle,qqwuqq); D((-1.2728718660159388,0.9649323513481336)--(-1.2578042173640724,1.0720604853321944)--(-1.3649323513481335,1.087128133984061)--B--cycle,qqwuqq); D(B--C); D(C--A); D(A--B); D((xmin,-0.14065071509700924*xmin+2.7946775614346366)--(xmax,-0.14065071509700924*xmax+2.7946775614346366)); D((xmin,-0.14065071509700924*xmin+0.7859020131661273)--(xmax,-0.14065071509700924*xmax+0.7859020131661273)); D((xmin,1.974358610218526*xmin+4.41601586521183)--(xmax,1.974358610218526*xmax+4.41601586521183)); D((xmin,-1.4710084190794759*xmin+0.1386871799331433)--(xmax,-1.4710084190794759*xmax+0.1386871799331433)); D(F--(-0.7665868345688304,2.9024985479006955)); D(M--C); D(A--D); D(D--B); D(A); MP("A",(-1.160593990386801,3.0079836849073933),NE*lsf); D(B); MP("B",(-1.425782334050844,0.896676487274441),NE*lsf); D(C); MP("C",(0.45603495079592204,2.1716204471977214),NE*lsf); D(M); MP("M",(-1.3441859206157538,1.9625296377703034),NE*lsf); D((-0.7665868345688304,2.9024985479006955)); MP("E",(-0.8036096816082817,2.9671854781898483),NE*lsf); D(F); MP("F",(-0.45172514866945557,0.896676487274441),NE*lsf); D(D); MP("D",(-0.6302173030587153,2.0747247062435523),NE*lsf); D(G); MP("G",(-0.9362038534403032,1.355656312846822),NE*lsf); D(H); MP("H",(-0.9209045259212237,2.778493772121203),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] After a nice diagram is drawn, it is evident that we not only want $\angle ADB = \angle EMF$; we also want $\triangle ADB \sim \triangle EMF \iff \angle MBD=\angle MFD, \angle MAD=\angle MED \iff (AMDE, BMDF$ are cyclic$) \iff MDF=90^\circ \iff \triangle MDF \sim \triangle MGC \iff \angle MFE = \angle MCB$. Let $G=MF \cap BC, H=ME \cap AC$. Then $\triangle AMH \sim \triangle AEM, \triangle BMG \sim \triangle BFM \Rightarrow ME/MA=MA/MH, MF/MB=MB/MG$. Since $MA=MC$, $(MH)(ME)=(MG)(MF) \Rightarrow (EF, GH$ are antiparallel WRT $M)\Rightarrow \angle MFE=\angle MHG=\angle MCB$ because $MHCG$ is cyclic, as desired.

Dear Mathlinkers, this nice problem is based on the concept of orthopole, then on the Pappus theorem and on a converse of the Reim's theorem. Sincerely Jean-Louis

My solution : Let $P$ and $Q$ the intersection of $ME$ and $AC$ and $MF$ and $BC$. We draw the parallel from $C$ to $AB$ cuts which $AE$ and $BF$ in $X$ and $Y$ respectively. Is easy too see that : $PEXC$ and $QFYC$ are cyclic, also $MA^{2}=MP.ME=MB^{2}=MQ.MF$ then $M$ It belongs to the radical axis from $PEXC$ and $QFYC$ then $D$ It belongs to the radical axis also is collinear with $E$ and $F$ Then $CM$ is perpendicular to $EF$. Also that $MA^{2}=MD.MC$ and $MB^{2}=MD.MC$ then $\angle MAD=\angle ACM$ and $\angle MCB=\angle BCD$. Then $\angle ADB=\angle EMF=180-\angle ACB$. Done!!

My solution: Take $D1$ on $MC$ duch that $MD1MC=MA^2$ now we have $\triangle AMD1 similar to\triangle AMC$ now we have $\angle AD1M=alpha=\angle AEM$ now $AEMD1$ is cyclic so$ \angle MD1E=90$ and analogously for $B$ and we get $E,F,D1$ colinear so $D1=D$ and its done.

http://www.artofproblemsolving.com/community/c6h598060p3548938 This kills it.

Let $\{X\}=ME\cap AC,\ \{Y\}=MF\cap BC$. Then $MX\cdot ME=MA^2=MB^2=MY\cdot MF$, hence $XYFE$ is cyclic. As $MXCY$ is cyclic, we get that $\widehat{XEF}=\widehat{MYX}=\widehat{MCA}$, i.e. $XDCE$ cyclic, hence $EF\perp CM$, which yields that $MDFB$ and $MDEA$ are cyclic. Now, the conclusion is obvious: $\widehat{MDB}=\widehat{MFB}=\hat{B}, \widehat{MDA}=\widehat{MEA}=\hat{A}$ so $\widehat{ADB}=\hat{B}+\hat{A}=\widehat{EMF}$.

Consider a coordinate system with origin $M(0,0)$. Now choose parameters $a,b$ and $c$ such that $A(-a,0)$, $B(a,0)$ and $C(b,c)$. Thus, the slope of $MC$ is $\frac{b}{c}$. Now the lines $BC$ and $CA$ have slopes $\frac{c}{b-a}$ and $\frac{c}{b+a}$, respectively. Therefore, the equation of the line $MF$ is $y=\frac{a-b}{c}x$ and the equation of $ME$ is $y=-\frac{a+b}{c}x$. Next, by using the fact that the equations of $AE$ and $BF$ are $x=-a$ and $x=a$ ,and by solving two systems of linear equations we get that $E(-a,\frac{a^2+ab}{c})$ and $F(a,\frac{a^2-ab}{c})$. Finally, the slope of $EF$ is $\frac{\frac{a^2-ab}{c}-\frac{a^2+ab}{c}}{a-(-a)}=-\frac{b}{c}$. Since the slopes of $MC$ and $EF$ are negative reciprocals, we get that the segments are perpendicular. Afterwards, we are done by observing the cyclic quadrilaterals $AMDE$ and $MBDF$.

I will share my solution even if it is a little bit "tangled" than yours. Let's denote the intersection between circumcircle of triangles $AEM$ and $BMF$ with $D^\prime $. Let $X$ and $Y$ be the intersection points of these circles with th esides $AC$ and $BC$, respectively. We want to prove that $D$ and $D^\prime $ represent the same point so it is enough to prove that $C$ is on the radical axis of the two circles. We will show that : $AC \cdot CX=BC \cdot CY$, which is nearly obivious using cosine theorem...

The proposer of that problem is the same with the proposer of this year's BMO geometry problem, Theoklitos Paragyiou from Cyprus.

Inversion also kills this problem!

Where is going to be held JBMO 2016?

JBMO 2016 is going to be hold in Slatina,Romania

Arslan wrote: Where is going to be held JBMO 2016? Romania slatina

Nice problem,..., inversion with center at M and radius c/2.

my simple solution: AE^2-ME^2=AE^2-EA^2-MA2=CM^2-2AM^2=FC^2-MF^2==> CM pedencular EF blah blah

Let ME meet AC at X and MF meet BC at Y. $MA^2$ = $MX\cdot ME$ and $MB^2$ = $MY\cdot MF$ But MA = MB. So, $MX\cdot ME$ = $MY\cdot MF$ , or EXYF is cyclic. So, MFE = MXY = MCB. Thus, DYFC is also cyclic. So, EF is perpendicular to MC. Hence, MDFB is cyclic. Similarly, AMDE is also cyclic. Using some simple angle chasing, we get the desired result.

Solution: Let $\angle ACM = \theta$. By Law of sines in triangles $ACM$ and $BCM$, we get \[\frac{\sin A}{\sin B}=\frac{\sin \theta}{\sin (C-\theta)}.\]Let $D'$ and $D''$ be the feet perpendicular from $E$ and $F$ respectively on $CM$. \[MD' = EM \sin \theta = \frac{c}{2\sin A}\cdot \sin \theta=\frac{c}{2\sin B}\cdot \sin(C-\theta)=FM\sin(C-\theta)=MD''.\]So, $D'=D'' = D$. So, $\angle EDM=\angle FDM=90^{\circ}, \Rightarrow EDMA, FDMB$ are cyclic. \[\angle EMF=180-C. \qquad \angle ADB=\angle MDA+\angle MDB=A+B=180-C.\] Proved.$\blacksquare$

I've renamed the points. neverlose wrote: Let $ABC$ be an acute triangle.The lines $l_1$ and $l_2$ are perpendicular to $BC$ at the points $B$ and $C$, respectively.The perpendicular lines from the midpoint $M$ of $BC$ to the lines $AB$ and $AC$ intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively.If $D$ is the intersection point of the lines $EF$ and $MA$, prove that \[\angle CDB = \angle EMF.\] We claim that $AM\perp EF$. From repeated application of Pythagoras theorem, $BE^2-BM^2+CM^2-CF^2+MF^2-ME^2 = CM^2-BM^2 = 0$. So, $MD\perp EF$ by carnot. Now from cyclic quadrilaterals $BEDM, CFDM$ we get $\angle BDC = \pi - \angle A$. Now we see that $\angle EMF = \pi -\angle A$, from right triangles $BME,CMF$. So the conclusion. $\blacksquare$

Lemma: Let $A,B$ be on $l_1$ and $C,D$ be on $l_2$. $l_1\perp l_2$ iff $AC^2+BD^2=AD^2+BC^2.$ Proof: Let $X=AB\cap CD$ and $\angle CXA=\alpha.$ The if part is trivial. For the only if part, note that \begin{align*} &AC^2=AX^2+CX^2-2(AX)(CX)\cos\alpha \\ &BD^2=DX^2+BX^2-2(DX)(BX)\cos\alpha \\ &AD^2=AX^2+DX^2+2(AX)(DX)\cos\alpha \\ &BC^2=CX^2+BX^2+2(CX)(BX)\cos\alpha. \end{align*}Substituting these values in $AC^2+BD^2=AD^2+BC^2$ gives $2\cos\alpha[(AX)(CX)+(CX)(BX)+(BX)(DX)+(DX)(AX)]=0 \implies \cos\alpha=0 \implies l_1\perp l_2.$ $\Box$ Since $AC\perp ME$ and $MF\perp BC$,by the Lemma, \begin{align} AM^2+EC^2=AE^2+CM^2\\ BM^2+CF^2=FB^2+MC^2. \end{align}. Subtracting (2) from (1) and noting that $MA^2=MB^2$, we get $MF^2+EC^2=ME^2+CF^2$. By the Lemma, this implies that $EM\perp MC$ so $EAMD$ and $FBMD$ are cyclic so $\angle DEM=\angle DAM$ and $\angle DFM=\angle DBM.$ Hence, $$\angle ADM=180^{\circ}-\angle DAB-\angle DBA=180^{\circ}-\angle MEF-\angle MFE=\angle EMF$$, as desired. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -9.858, xmax = 9.862, ymin = -5.608, ymax = 5.612; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.08,-1.21)--(2.74,-1.21)--(-2.348,4.412)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.08,-1.21)--(2.74,-1.21), linewidth(2) + rvwvcq); draw((2.74,-1.21)--(-2.348,4.412), linewidth(2) + rvwvcq); draw((-2.348,4.412)--(-4.08,-1.21), linewidth(2) + rvwvcq); draw(circle((-0.67,0.8172561366061897), 3.967098113660344), linewidth(2) + wrwrwr); draw((-4.08,ymin)--(-4.08,ymax), linewidth(2) + wrwrwr); /* line */ draw((2.74,ymin)--(2.74,ymax), linewidth(2) + wrwrwr); /* line */ draw((2.74,-2.2605371753824257)--(-4.08,-4.296104589114194), linewidth(2) + wrwrwr); draw((2.74,1.876104589114194)--(-4.08,-0.15946282461757333), linewidth(2) + wrwrwr); draw((-4.08,-0.15946282461757333)--(2.74,-2.2605371753824257), linewidth(2) + wrwrwr); draw((-0.67,-1.21)--(2.74,1.876104589114194), linewidth(2) + wrwrwr); draw((-0.67,-1.21)--(-4.08,-4.296104589114194), linewidth(2) + wrwrwr); draw((-2.348,4.412)--(-0.10316396963759333,-3.1091371648971693), linewidth(2) + wrwrwr); /* dots and labels */ dot((-4.08,-1.21),dotstyle); label("$A$", (-3.998,-1.008), NE * labelscalefactor); dot((2.74,-1.21),dotstyle); label("$B$", (2.822,-1.008), NE * labelscalefactor); dot((-2.348,4.412),dotstyle); label("$C$", (-2.258,4.612), NE * labelscalefactor); dot((-0.67,-1.21),linewidth(4pt) + dotstyle); label("$M$", (-0.598,-1.048), NE * labelscalefactor); dot((2.74,-2.2605371753824257),linewidth(4pt) + dotstyle); label("$Y$", (2.822,-2.108), NE * labelscalefactor); dot((-0.10316396963759333,-3.1091371648971693),linewidth(4pt) + dotstyle); label("$Z$", (-0.018,-2.948), NE * labelscalefactor); dot((-4.08,-4.296104589114194),linewidth(4pt) + dotstyle); label("$X$", (-3.998,-4.128), NE * labelscalefactor); dot((-4.08,-0.15946282461757333),linewidth(4pt) + dotstyle); label("E", (-3.998,-0.008), NE * labelscalefactor); dot((2.74,1.876104589114194),linewidth(4pt) + dotstyle); label("F", (2.822,2.032), NE * labelscalefactor); dot((-1.2368360303624062,0.6891371648971687),linewidth(4pt) + dotstyle); label("D", (-1.158,0.852), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]Let the reflections of $D, E, F$ over $M$ be $Z, Y, X$. It suffices to show that $\angle AZB = \angle EMF = 180^\circ - \angle ACB$, or that $ABCZ$ is cyclic. Let us redefine $Z$ as the second intersection of $CM$ and $\Gamma$, the circumcircle of $\Delta ABC$, and we will show that $X, Y, Z$ are collinear. Put the figure on the cartesian plane with $A = (-1,0), B = (1, 0), C = (x, y)$. Note $\angle YMB = \angle CAB$, so by similar triangles we have $Y = (1,-\frac{x+1}{y})$ and $X = (-1, \frac{x-1}{y})$. By power of a point we have $MZ = \frac{1}{CM} = \frac{\sqrt{x^2+y^2}}{x^2+y^2}$; since $C, M, Z$ are collinear we have $Z = (-\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2})$. Now we just need to show $Z, X, Y$ are collinear, or that $$\frac{\frac{x}{x^2+y^2}-1}{\frac{y}{x^2+y^2}+\frac{x-1}{y}} = \frac{\frac{x}{x^2+y^2}+1}{\frac{y}{x^2+y^2}-\frac{x+1}{y}}$$, which is true by expansion.

[asy][asy] import markers; unitsize(1inch); pair A, B, C, D, E, F, M, G, I, X, Y, P; A = dir(200); B = dir(340); C = dir(110); M = .5A + .5B; G = foot(M, A, C); I = foot(M, B, C); X = dir(160); Y = dir(20); E = extension(M, G, A, X); F = extension(M, I, B, Y); D = extension(C, M, E, F); draw(A--B--C--cycle, lightblue); draw(E--F, lightblue); draw(A--E); draw(B--F); draw(C--M, lightblue); P = foot(A, B, C); draw(arc(M, B, A), dashed+heavygreen); draw(circumcircle(M, C, I), dotted); draw(E--M--F, deepmagenta); draw(A--D--B, deepmagenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)+1); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", G, dir(30)+.3); dot("$Y$", I, dir(0)+.4); markangle(n=1, radius = 11.25, M, A, C); markangle(n=1, radius = 11.25, A, D, M); markangle(n=2, radius = 10, M, D, B); markangle(n=2, radius = 10, C, B, M); [/asy][/asy] Let $X, Y$ be the feet from $M$ to $CA, CB$. Denote by $\omega$ the circle with center $M$ and radius $MA$. Since $EX\cdot EM = EA^2$ and $FY\cdot FM = FB^2$, $EF$ is the radical axis of $\omega$ and $(CXMY)$. The power of $D$ wrt $\omega$ is $MA^2 - MD^2$, and the power of $D$ wrt $(CXMY)$ is $CD\cdot MD$, so \[MA^2 - MD^2 = CD\cdot MD\implies MA^2 = MD\cdot MC.\]Then $\triangle MDA\sim\triangle MAC$ and $\triangle MDB\sim\triangle MBC$, so \[\angle ADB = \angle MDA + \angle MDB = \angle A + \angle B = \angle EMF.\]

This is so lame compared to everyone else's but just because it wasn't mentioned before... Let $A= a$, $B= b$, and $C=c$ such that $A, B, C$ lie on the unit circle. It can be shown that $E = e = \frac{3ac-bc-ab-b^2}{2(c-b)}$ and $F$ the same but variables $a$ and $b$ flipped. It can be shown easily from here that $EF$ and $CM$ are perpendicular.

anser wrote: [asy][asy] import markers; unitsize(1inch); pair A, B, C, D, E, F, M, G, I, X, Y, P; A = dir(200); B = dir(340); C = dir(110); M = .5A + .5B; G = foot(M, A, C); I = foot(M, B, C); X = dir(160); Y = dir(20); E = extension(M, G, A, X); F = extension(M, I, B, Y); D = extension(C, M, E, F); draw(A--B--C--cycle, lightblue); draw(E--F, lightblue); draw(A--E); draw(B--F); draw(C--M, lightblue); P = foot(A, B, C); draw(arc(M, B, A), dashed+heavygreen); draw(circumcircle(M, C, I), dotted); draw(E--M--F, deepmagenta); draw(A--D--B, deepmagenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)+1); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); dot("$X$", G, dir(30)+.3); dot("$Y$", I, dir(0)+.4); markangle(n=1, radius = 11.25, M, A, C); markangle(n=1, radius = 11.25, A, D, M); markangle(n=2, radius = 10, M, D, B); markangle(n=2, radius = 10, C, B, M); [/asy][/asy] Let $X, Y$ be the feet from $M$ to $CA, CB$. Denote by $\omega$ the circle with center $M$ and radius $MA$. Since $EX\cdot EM = EA^2$ and $FY\cdot FM = FB^2$, $EF$ is the radical axis of $\omega$ and $(CXMY)$. The power of $D$ wrt $\omega$ is $MA^2 - MD^2$, and the power of $D$ wrt $(CXMY)$ is $CD\cdot MD$, so \[MA^2 - MD^2 = CD\cdot MD\implies MA^2 = MD\cdot MC.\]Then $\triangle MDA\sim\triangle MAC$ and $\triangle MDB\sim\triangle MBC$, so \[\angle ADB = \angle MDA + \angle MDB = \angle A + \angle B = \angle EMF.\] Super cool!!

Let $ME$ intersects $AC$ at point $G$, let $MF$ intersects $BC$ at $H$. Triangles $MGA$ and $MAE$ are similar, so: $\frac{MA}{MG}=\frac{ME}{MA} \implies MA^2=MG \cdot ME$ Triangles $MBH$ and $MFB$ are similar, so: $\frac{MB}{MH}=\frac{MF}{MB} \implies MB^2=MF \cdot MH$ This means that $MF \cdot MH=MG \cdot ME$, so triangles $MEF$ and $MGH$ are similar, $\angle MGH=\angle EFG=180^\circ-\angle EGH$ From this, quadrilateral $EGHF$ is cyclic. Also, $\angle CDE=\angle EDM=90^\circ$. This means that we have more cyclic quadrilaterals $FDMB, EAMD$, because $\angle MBF+\angle MDF=180^\circ$ and $\angle MAE+\angle EDM=180^\circ$, so $\angle DAB=\angle FEM, \angle DBA=\angle EFM \implies EMF$ and $ADB$ are similar, so $\angle ADB=\angle EMF$.

Let the line $ME$ intersect $BF$ at $H$, and let $R$ be the midpoint of $AC$ (so $RM \parallel BC$). Then $\angle FHM = 90 - \angle HMB = 90 - \angle EMA = \angle MAR$. Also, $\angle MFH = \angle MFB = \angle CBM = \angle RMA$. Thus, $\triangle FHM \sim \triangle MAR$. Since $ME=MH$ and $AR=RC$, we in fact have $\triangle FHE \sim \triangle MAC$. Then angle between corresponding sides are equal, so the angle between $EF$ and $CM$ is the angle between $EH$ and $CA$, which is $90$. Thus, $EDMA$ and $FDMB$ are both cyclic as they have two opposite right angles. Then $\angle ADB = \angle ADM + \angle BDM = \angle AEM + \angle BFM = (90- \angle AME) + (90- \angle FMB) = \angle EMF$.

Let $J=AC\cap ME$ and $I=BC\cap MF$. Then we have $MA^2 = MJ\cdot ME = MB^2 = MI\cdot MF \implies FIJE$ is cyclic. So we have $\angle JEF = \angle JIM = \angle JCM \implies EJDC$ is cyclic $\implies \angle EDC = 90 \implies AMDE$ and $BMDF$ are cyclic. So we have $\angle ADB = \angle MDB + \angle MDA = \angle MFB + \angle MEA = (90 - \angle AME) + (90 - \angle BMF) = 180 -(\angle AME + \angle BMF) = \angle EMF$. We are done.

The problems that the author of this problem proposes are just amazing !

nois

Let MF meet BC at H and ME meet AC at G. MG.ME = MA^2 = MB^2 = MH.MF ---> HFEG is cyclic. ∠EFH = ∠MGH = ∠MCH and ∠MCH + ∠CMH = 90 ---> MDF = 90 ---> MDFB is cyclic. MDEA will be cyclic the same way so: ∠EMF = ∠AEM + ∠MFB = ∠ADM + ∠BDM = ∠ADB so we're Done.

I believe no one has stated the obvious fact about $D$ yet. Interesting. Let $X$ and $Y$ be the intersection points of $ME$ and $MF$ with $AC$ and $BC$ respectively. Claim. $X,E,Y,F$ are concyclic. Proof. Upon applying Euclid's Theorem to $\triangle{MAE}$ and $\triangle{MBF}$, we obtain $$MX\cdot ME=MA^2=MB^2=MY\cdot MF$$which implies the claim. $\blacksquare$ Claim. $D$ is the $C$-Humpty Point. Proof. It is trivial that $M,X,C,Y$ are concyclic. Chasing angles using the previous claim and this fact we get that $$\angle{XCD}=\angle{MAX}=\angle{MYX}=\angle{XEY}=\angle{XED}$$holds. This implies $X,C,E,D$ are concyclic. Using PoP from $M$ gives $$MA^2=MX\cdot ME=MD\cdot MC$$which implies that $(CDA)$ is tangent to $AB$. From symmetry, $(CDB)$ is also tangent to $AB$. Hence, the claim is proven since this is enough to imply that $D$ is the Humpty Point. $\blacksquare$ Because $D$ is the $C$-Humpty Point, $A,H,D,B$ are concyclic. From this we obtain $$\angle{ADB}=\angle{AHB}=180-\angle{ACB}=180-\angle{XCY}=\angle{XMY}=\angle{EMF}$$which finishes the proof. $\blacksquare$

Easy i did it like anyone else with my friend Lyron but we did it in a better way.