My solution: Let's observe that $DE=AD\cdot\sin(\angle{BAD})$ and $DF=AD\cdot\sin(\angle{CAD})$,so $\frac{DE}{DF}=\frac{AB}{AC}$.We have $$m(\angle{EDM})=m(\angle{EDA})+m(\angle{ADM})=90^0-m(\angle{BAO})-m(\angle{OAD})+m(\angle{EDM})=90^0-m(\angle{BAO})=m(\angle{ACB})$$and $$m(\angle{MDF})=m(\angle{EDF})-m(\angle{EDM})=180^0-m(\angle{BAC})-m(\angle{ACB})=m(\angle{ABC})$$. This implies that $\frac{\sin(\angle{EDM})}{\sin(\angle{FDM})}=\frac{\sin(\angle{ACB})}{\sin(\angle{ABC})}=\frac{AB}{AC}=\frac{ED}{FD}$,so $EM=FM$.

My solution: Let $M$ be a midpoint of $EF$ we will prove that $AO\parallel DM$ Let $X$ the intersection of $AD$ and the circumcircle of $\triangle ABC$ $\Longrightarrow$ $\triangle BXC\sim \triangle EDF$ $\Longrightarrow$ $\angle DXC=\angle MDC=\angle ABC$ Let $Y=BC\cap AO$ $\Longrightarrow$ $\angle AYC=90^{\circ}-\angle ACB+\angle ABC$ and it is easy to see that $\angle MDC=90^{\circ}-\angle ACB+\angle ABC$ $\Longrightarrow$ $\angle AYC=\angle MDC$ $\Longrightarrow$ $AO\parallel DM$...

huricane wrote: My solution: Let's observe that $DE=AD\cdot\sin(\angle{BAM})$ and $DF=AD\cdot\sin(\angle{CAM})$,so $\frac{DE}{DF}=\frac{AB}{AC}$.We have $$m(\angle{EDM})=m(\angle{EDA})+m(\angle{ADM})=90^0-m(\angle{BAO})-m(\angle{OAD})+m(\angle{EDM})=90^0-m(\angle{BAO})=m(\angle{ACB})$$and $$m(\angle{MDF})=m(\angle{EDF})-m(\angle{EDM})=180^0-m(\angle{BAC})-m(\angle{ACB})=m(\angle{ABC})$$. This implies that $\frac{\sin(\angle{EDM})}{\sin(\angle{FDM})}=\frac{\sin(\angle{ACB})}{\sin(\angle{ABC})}=\frac{AB}{AC}=\frac{ED}{FD}$,so $EM=FM$. do you have a figure for this? Can you please share it?

here is my solution = by sine law in triangles $BAD,CAD$ and using $\angle DEF=\angle DAC$ and $\angle DFE=\angle DAE$ we get $\frac{AC}{AB}=\frac{DE}{DF}$ again using sine law in triangles $DME,DMF$ yields $\frac{DEsinEDM}{DFsinFDM}=\frac{ACsinEDM}{ABsinFDM}=\frac{ME}{MF}$. so it remains to show that $\frac{AC}{AB}=\frac{sinB}{sinC}=\frac{sinFDM}{sinEDM}$ which after some angle chasing follows as let $DM\cap AC=K$ than $\angle DKA+\angle OAK+\angle OAE+\angle AED+\angle EDM=180+90+90-\angle C+\angle EDM=360$ so that $\angle EDM=\angle C$. similarly $\angle FDM=\angle B$. so that $\frac{AC}{AB}=\frac{sinB}{sinC}=\frac{sinFDM}{sinEDM}$ holds true and hence $EM=MF$. so we are done

From $\angle DEM=\angle DAF$ and $\angle EDM=90^{\circ}-\angle OAE=\angle C$ we have $\triangle EDM\sim \triangle ACD$, similarly $\triangle FDM\sim \triangle ABD$. Then $\frac{EM}{DM}=\frac{AD}{CD}=\frac{AD}{BD}=\frac{FM}{DM}$ so $EM=FM$.

aditya21 wrote: which after some angle chasing follows as let $DM\cap AC=K$ than $\angle DKA+\angle OAK+\angle OAE+\angle AED+\angle EDM=180+90+90-\angle C+\angle EDM=360$ so that $\angle EDM=\angle C$. similarly $\angle FDM=\angle B$. how did this happen? I only understand $\angle DKA+\angle OAK=180$..

leeky wrote: $\angle EDM=90^{\circ}-\angle OAE=\angle C$ can you please explain this?

dftba123 wrote: aditya21 wrote: which after some angle chasing follows as let $DM\cap AC=K$ than $\angle DKA+\angle OAK+\angle OAE+\angle AED+\angle EDM=180+90+90-\angle C+\angle EDM=360$ so that $\angle EDM=\angle C$. similarly $\angle FDM=\angle B$. how did this happen? I only understand $\angle DKA+\angle OAK=180$.. since $AD$ is diameter of circle $ADE$ thus it follows that $\angle DEA=90$ also $\angle OAE=\angle OAB=90-\angle C$ which is direct consequence of $OA=OB$ combined with $\angle OAB=2\angle C$ and thus $\angle EDM=\angle C$ and similar process will give us $\angle FDM=\angle B$ i hope you understand it now.

aditya21 wrote: dftba123 wrote: aditya21 wrote: which after some angle chasing follows as let $DM\cap AC=K$ than $\angle DKA+\angle OAK+\angle OAE+\angle AED+\angle EDM=180+90+90-\angle C+\angle EDM=360$ so that $\angle EDM=\angle C$. similarly $\angle FDM=\angle B$. how did this happen? I only understand $\angle DKA+\angle OAK=180$.. since $AD$ is diameter of circle $ADE$ thus it follows that $\angle DEA=90$ also $\angle OAE=\angle OAB=90-\angle C$ which is direct consequence of $OA=OB$ combined with $\angle OAB=2\angle C$ and thus $\angle EDM=\angle C$ and similar process will give us $\angle FDM=\angle B$ i hope you understand it now. Yeah I do. Thanks!

Here's a diagram:

My solution: We know $\frac{ED}{DC}=\frac{Sin(ABC)\cdot Sin (BAD)}{Sin(DFE)},$ (Applying the Sine Law on $\triangle EBD$ and use $DB=DC.$) and $\frac{DF}{DC}=\frac{Sin(ACB)\cdot Sin(CAD)}{Sin(DEF)}.$ (Applying the Sine Law on $\triangle DFC$ and use $DB=DC.$) Then we can find easily $\frac{Sin(DEF)}{Sin(DFE)}=\frac{Sin(ABC)}{Sin(ACB)}.$ ( Use $\frac{Sin(BAD)}{Sin(CAD)}=\frac{Sin(ABC)}{Sin(ACB)}.$) $(1).$ We must to show that $$\frac{Sin(FDM)}{Sin(EDM)}=\frac{Sin(DFE)}{Sin(DEF)}=^{(1)}=\frac{Sin(ABC)}{Sin(ACB)}.$$Let $H$ be the orthocentre this triangle,and $K$ be the midpoint of $AH.$ Then we know $AODK$ is parallelogram.Let $\angle BAO=\angle CAK=z,$ and $\angle OAD=y,\angle KAD=x.$ Then we can find easily $\angle FDM=90-x-y-z,\angle EDM=90-z,\angle ABC=90-x-y-z$ and $\angle ACB=90-z.$ Then $\frac{Sin(FDM)}{Sin(EDM)}=\frac{Sin(90-x-y-z)}{Sin(90-z)}=\frac{Sin(ABC)}{Sin(ACB)}.$ As desired.

My solution: I'll use the diagram from Post #12. Let $DM \cap AC=K$. Then $$\angle DKC=\angle OAC=90^{\circ}-\angle ABC \Rightarrow \angle MDF=90^{\circ}-\angle DKF=\angle ABD$$Also, $\angle DFM=\angle DAB \Rightarrow \triangle FDM \sim \triangle ABD$. Similarly, we have $\triangle EDM \sim \triangle ACD$. Thus, Using the fact that $BD=CD$, we get that $\frac{EM}{MD}=\frac{AD}{CD}$ and $\frac{FM}{MD}=\frac{AD}{BD}$ $\Rightarrow EM=FM$ $\blacksquare$

Let $AT \perp EF$ be $A-$symmedian of $\angle A$ $AL$ is the tangent at $A$ of $(O)$ $DM \parallel AO \Rightarrow DM \perp AL$ We have: $-1=A(B,C;L,T)=D(E,F;M,\infty),$ so $M$ is the midpoint of $EF$

We instead prove that if $M$ is the midpoint of $EF$ then $DM || AO$. By a simple angle chase, this is equivalent to showing that $\angle EDM = C$ (or $\angle FDM = B$). [asy][asy] import olympiad;size(5cm);pair A,B,C;A=dir(110);B=dir(210);C=dir(330);draw(A--B--C--cycle);pair D = (B+C)/2;pair E = foot(D,A,B);pair F = foot(D,A,C);draw(D--E--F--D);draw(rightanglemark(B,E,D,2));draw(rightanglemark(D,F,C,2));pair M = (E+F)/2;pair H = (2*M-D);draw(A--D);draw(D--H,dotted);draw(E--H--F,dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(M)); dot("$H$",H,dir(H)); [/asy][/asy] Reflect $D$ over $M$ to get $H$. Then $\angle DAC = \angle DAF = \angle DEF = \angle DEM$, but also $\angle BAC = \angle EAD = \angle EFD = \angle HEF = \angle HEM$. Due to these equalities and the fact that $AD$ and $EM$ are the medians of $\triangle ABC$ and $\triangle EHD$ respectively, this is enough to conclude that $\triangle ABC \sim \triangle EHD$. It follows that $\angle C = \angle EDH$, done.

FabrizioFelen wrote: My solution: Let $M$ be a midpoint of $EF$ we will prove that $AO\parallel DM$ Let $X$ the intersection of $AD$ and the circumcircle of $\triangle ABC$ $\Longrightarrow$ $\triangle BXC\sim \triangle EDF$ $\Longrightarrow$ $\angle DXC=\angle MDC=\angle ABC$ Let $Y=BC\cap AO$ $\Longrightarrow$ $\angle AYC=90^{\circ}-\angle ACB+\angle ABC$ and it is easy to see that $\angle MDC=90^{\circ}-\angle ACB+\angle ABC$ $\Longrightarrow$ $\angle AYC=\angle MDC$ $\Longrightarrow$ $AO\parallel DM$... $\triangle BXC\sim \triangle EDF$ isn't necessarily true, is it?

huricane wrote: My solution: Let's observe that $DE=AD\cdot\sin(\angle{BAM})$ and $DF=AD\cdot\sin(\angle{CAM})$,so $\frac{DE}{DF}=\frac{AB}{AC}$.We have $$m(\angle{EDM})=m(\angle{EDA})+m(\angle{ADM})=90^0-m(\angle{BAO})-m(\angle{OAD})+m(\angle{EDM})=90^0-m(\angle{BAO})=m(\angle{ACB})$$and $$m(\angle{MDF})=m(\angle{EDF})-m(\angle{EDM})=180^0-m(\angle{BAC})-m(\angle{ACB})=m(\angle{ABC})$$. This implies that $\frac{\sin(\angle{EDM})}{\sin(\angle{FDM})}=\frac{\sin(\angle{ACB})}{\sin(\angle{ABC})}=\frac{AB}{AC}=\frac{ED}{FD}$,so $EM=FM$. Is the first two lines always true? Or should it be $\angle{BAD}$ and $\angle{CAD}$ since $A$, $M$, and $D$ are not necessarily collinear?

@above It should be fixed now.

Why so convoluted? Note that $\angle EDM=\angle C,\angle FDM=\angle B$. Hence $\dfrac{EM}{FM}=\dfrac{DE\sin C}{DF\sin B}=\dfrac{BD\sin B \sin C}{DC\sin B \sin C}=1\blacksquare$

So first of all we must calculate some angles. $$\angle OAB = 90 - \angle C = \angle TAB \implies \angle CTA = 90-\angle C + \angle B$$$$\angle CTA = \angle CDM = 90-\angle C + \angle B = \angle CDF + \angle FDM \implies \angle FDM = \angle B$$$$\angle MDE = 180-\angle A - \angle B = \angle C$$So now let's start with the trigonometry: We know that $MF = \frac{sin \angle B}{sin \angle DFE}MD$,but we also know that $ME=\frac{sin \angle C}{sin \angle DEF}MD$ From here we know that $ME = MF$, because we have that $\frac{sin \angle B}{sin \angle DFE} = \frac{sin \angle B}{sin \angle DAE} = \frac{AD}{BD} = \frac{AD}{BC} = \frac{sin \angle C}{sin \angle FAD} =\frac{sin \angle C}{sin \angle DEF} $

Let's start by adding the heights: $BP$ and $CQ$. Let $X$ be the midpoint of $PQ$. It is obvious that $EDFX$ is a parallelogram and also that $DX \perp PQ$ and since $AO \perp PQ$ we get $DX \parallel AO$ and we know $(D,M,X)$ are collinear hence the claim.

∠DEM = ∠DEF = ∠DAC and ∠MDE = 90 - ∠OAB = ∠C so EMD and ACD are similar. with same approach we can prove FDM and ABD are similar as well. EM/AD = MD/DC = MD/BD = FM/AD so EM = FM. we're Done.

Notice $$\measuredangle MDE=\measuredangle(\overline{AO},\overline{DE})=90-\measuredangle BAO=\measuredangle ACB$$and similarly $\measuredangle FDM=\measuredangle CBA.$ By the Ratio Lemma, $$\frac{EM}{FM}=\frac{ED\cdot\sin\angle MDE}{FD\cdot\sin\angle FDM}=\frac{ED\cdot\sin\angle C}{FD\cdot\sin\angle B}=\frac{DB}{DC}=1.$$$\square$

Pluto1708 wrote: Why so convoluted? Note that $\angle EDM=\angle C,\angle FDM=\angle B$. Hence $\dfrac{EM}{FM}=\dfrac{DE\sin C}{DF\sin B}=\dfrac{BD\sin B \sin C}{DC\sin B \sin C}=1\blacksquare$ I should use Trig more often math_pi_rate wrote: My solution: I'll use the diagram from Post #12. Let $DM \cap AC=K$. Then $$\angle DKC=\angle OAC=90^{\circ}-\angle ABC \Rightarrow \angle MDF=90^{\circ}-\angle DKF=\angle ABD$$Also, $\angle DFM=\angle DAB \Rightarrow \triangle FDM \sim \triangle ABD$. Similarly, we have $\triangle EDM \sim \triangle ACD$. Thus, Using the fact that $BD=CD$, we get that $\frac{EM}{MD}=\frac{AD}{CD}$ and $\frac{FM}{MD}=\frac{AD}{BD}$ $\Rightarrow EM=FM$ $\blacksquare$ The founder of the Q point himself didn’t use it….. strange. Let $\odot(AD) \cap \odot(ABC)=Q$ And $AO \cap \odot(ABC)=A’$ Easy to see that $Q$ is the Queue point and hence $Q-D-A’$ $$\measuredangle QDM= \measuredangle QA’A = \measuredangle QBA= \measuredangle QBE$$We also know that this $Q$ is the (Miquel pt of $BCFE$) center of spiral similarity taking $EB \longrightarrow FC$ which combined with the above statement means Q takes $EB \longrightarrow MD \implies EM=MF \ \blacksquare$ Well… this solution is also not long…. Not that bad huh

$AO\cap EF=K,AO\cap ED=N,AO\cap FD=L$ $\angle EDK=90-\angle BDE=\angle B$ and $\angle KDF=90-\angle FDC=\angle C$ $\angle FAL=90-\angle B \implies MDF=\angle ALF=\angle B \implies \angle KDM=\angle C-\angle B$ $ED=\sin B.\frac{BC}{2}$ and $DF=\sin C.\frac{BC}{2} \implies \frac{ED^2}{DF^2}=\frac{\sin^2 B}{\sin^2 C}$ $EK=\frac{KD}{\sin FED}.\sin B=\frac{KD}{\sin FAD}.\sin B$ and $KF=\frac{KD}{\sin DFE}.\sin C=\frac{KD}{\sin DAE}.\sin C \implies \frac{EK}{KF}=\frac{\sin DAE}{sin FAD}.\frac{\sin B}{\sin C}$ Also $\sin DAE=\sin DAB=\frac{BD}{AD}.\sin B$ and $\sin FAD=\sin CAD=\frac{CD}{AD}.\sin C \implies \frac{\sin DAE}{\sin FAD}=\frac{\sin B}{\sin C}$ We get $\frac{EK}{KF}=\frac{\sin DAE}{sin FAD}.\frac{\sin B}{\sin C}=\frac{\sin^2 B}{\sin^2 C}=\frac{ED^2}{DF^2}$ and $\angle EDK=\angle B=\angle MDF$ gives us that $DK$ is $D-$symmedian in triangle $DEF$ and $DM$ is the median as desired.$\blacksquare$

Denote by $M$ the midpoint of segment $EF$ and by $H_b$ and $H_c$ the feet of the altitudes from $B$ and $C$. Let $H$ denote the orthocenter of $\triangle ABC$ and $N_a$ the midpoint of segment $AH$. We first remind ourselves the following well known result. Claim : Quadrilateral $N_aDOA$ is a parallelogram. Proof : Let $A'$ denote the $A-$antipodal point in $(ABC)$. It is well known that points $H$ , $D$ and $A'$ are collinear. Note that since $OM \perp BC \perp AH$ , $AH \parallel OD$. Further we know that $HM=MA'$ so the homothety centered at $A'$ with scale factor $2$ maps $OD$ to $AH$. Thus, \[2AN_a = AH = 2OD\]so $AN_a=OD$. Since we noted that these lines are also clearly parallel, it follows that $N_aDOA$ is a parallelogram as claimed. Now, note that by Newton-Gauss on self-intersecting quadrilateral $BH_bCH_c$, points $D$ , $N_a$ and $L$ - the midpoint of segment $H_bH_c$ are collinear. To finish, we note the following observation. Claim : Points $E$ and $F$ are respectively the midpoints of segments $BH_c$ and $CH_b$. Proof : Note that since $DE \perp AB \perp CH_c$, lines $CH_c$ and $DE$ are parallel. Since $D$ is the midpoint of segment $BC$, it thus follows that $E$ is also the midpoint of segment $BH_c$. The proof of the other is entirely similar. We finish by noting that since $E$ is the midpoint of $H_cB$ and $F$ is the midpoint of $H_bC$, by the Mean Geometry Theorem the midpoints of segments $H_bH_c$ , $EF$ and $BC$ are also collinear. Thus, points $L$ , $M$ and $D$ are collinear. Since we already noted that points $N_a$ , $L$ and $D$ are collinear, this means that $\overline{DM}$ intersects segment $AH$ at $N_a$. As a result of our first claim it follows that $DM \parallel AO$ and we are done.