Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
Problem
Source: CGMO 2007 P5
Tags: geometry, circumcircle, geometric transformation, reflection, homothety, parallelogram, Hi
28.12.2008 14:30
April wrote: Point $ D$ lies inside triangle $ ABC$ such that $ \angle DAC = \angle DCA = 30^{\circ}$ and $ \angle DBA = 60^{\circ}$. Point $ E$ is the midpoint of segment $ BC$. Point $ F$ lies on segment $ AC$ with $ AF = 2FC$. Prove that $ DE \perp EF$.
29.12.2008 01:24
Let $ (K)$ be, the circumcircle of the triangle $ \bigtriangleup ABD,$ with $ \angle ABD = 60^{o}.$ From the isosceles triangle $ \bigtriangleup DAC,$ with $ \angle DAC = \angle DCA = 30^{o}$ and $ AF = 2FC,$ it is easy to show that $ DF = FC$ and so, we have that $ \angle DFA = 60^{o}$ $ \Longrightarrow$ $ AD\perp DF.$ We denote the point $ B'\equiv (K)\cap DF$ and let $ K'$ be, the midpoint of the segment $ DF.$ Applying the Menelaos theorem in the equilateral triangle $ \bigtriangleup AB'F,$ we can say that the points $ K,\ K',\ C$ are collinear, because of $ \frac {KA}{KB'}\cdot \frac {K'B'}{K'F}\cdot \frac {CF}{CA} = 1$ $ ,(1)$ where $ K$ is the center of $ (K).$ $ ($ from $ AD\perp B'DF$ we have that $ AB'$ is a diameter of $ (K)$ and from $ AB' = B'F$ $ \Longrightarrow$ $ K'B' = 3K'F$ $ ).$ If we consider now, the triangle $ \bigtriangleup KAC,$ taken as its transversal the line segment $ B'F,$ applying again the Menelaos theorem, we have that $ \frac {K'K}{K'C}\cdot \frac {FC}{FA}\cdot \frac {B'A}{B'K} = 1$ $ \Longrightarrow$ $ \frac {K'K}{K'C} = 1$ $ \Longrightarrow$ $ K'K = K'C$ $ ,(2)$ $ \bullet$ From $ (2)$ and because of $ EB = EC,$ we conclude that $ K'E\parallel KB$ and $ KB = 2K'E$ $ ,(3)$ From $ AB' = B'F$ $ \Longrightarrow$ $ AK = DF$ $ \Longrightarrow$ $ KB = 2K'F$ $ ,(4)$ From $ (3),$ $ (4)$ $ \Longrightarrow$ $ K'E = K'F = K'D$ $ ,(5)$ From $ (5)$ we conclude that $ DF\perp EF$ and the proof is completed. Kostas Vittas.
Attachments:
t=247620.pdf (6kb)
02.01.2009 01:21
If $ M$ is midpoint of $ AC$ and $ G$ reflection of $ F$ in $ DM,$ then $ \frac{AC}{MC} = 2 = \frac{GC}{FC}.$ Since $ \frac{DC}{DM} = 2 = \frac{FM}{FC},$ $ DF$ bisects $ \angle CDM.$ $ \triangle DFG$ is therefore equilateral and $ \angle DGA = 120^\circ = 180^\circ - \angle DBA$ $ \Longrightarrow$ $ B$ is on circumcircle $ (P)$ of the isosceles $ \triangle DGA.$ Let $ CD$ cut $ (P)$ again at $ K.$ $ \triangle DAK$ is equilateral with $ \angle DKA = \angle ADK = 60^\circ$ $ \Longrightarrow$ $ \frac{KC}{DC} = 2.$ Let $ (Q)$ be circumcircle with diameter $ DF$ of the right $ \triangle DFM.$ Since $ \frac{KC}{DC} = \frac{AC}{MC} = \frac{GC}{FC} = 2,$ the circles $ (P), (Q)$ are similar with center $ C$ and coefficient 2. Since $ B \in (P)$ and $ \frac{BC}{EC} = 2,$ it follows that $ E \in (Q)$ with diameter $ DF$ and $ \angle DEF = 90^\circ.$
28.10.2010 03:23
Take $G$ on $AC$ so that $2AG = GC$. It is easy to see that $\angle GDA = \angle GAD = 30^{\circ}$. Let $D'$ be the reflection of $C$ across $D$. $D'DA = 180^{\circ} - \angle ADC = 60^{\circ}$, and $DA = DD'$, so $\triangle AD'D$ is equilateral, so $AD'BD$ is cyclic. Since $\angle AGD = 30^{\circ}$, $\angle GDD' = \angle GAD' = 90^{\circ}$, so $AGDBD'$ is cyclic. $\angle GD'D = \angle GAD = 30^{\circ}$ and $\angle D'AE = 60^{\circ}$, so $AD \perp D'G$, so $D'G$ must be a diameter of $D'AGDB$, so $\angle GBD' = 90^{\circ}$. Since $CG = 2CF$, a homothety centered at $C$ with factor 1/2 gives $FE \perp DE$, as desired.
13.11.2012 05:59
Solution: Assume that $CD$ meets $AB$ at $W$. Let $F'$ be the mid-point of $AF$ and $M$ is that of $AC$. Now apply cosine rules to show $DFF'$ is an equilateral triangle. So $\angle DBA=\angle DF'F=60^{\circ}$ and $ABDF'$ is cyclic. Since $\angle A+\angle B+\angle C=180^{\circ}$, we must have $\angle DAB+\angle DBC+\angle DCB=60^{\circ}$ In other words $\angle WDB+\angle DAB=60^{\circ}$ Hence \[\angle EFD=\angle EFA-60^{\circ}=\angle BF'A-60^{\circ} =\angle BDA-60^{\circ}=\angle BDW=60^{\circ}-\angle DAB\]At the same time $\angle EMD=\angle DXA\; (X=MD\cap AB)\; =90^{\circ}-\angle A=60^{\circ}-\angle DAB$ Therefore $\angle EFD=\angle EMD$ and $MDEF$ is cyclic. So $\angle DEF=180^{\circ}-\angle DMF=90^{\circ}$
10.04.2013 01:51
Without loss of generality, $AC = 3$. Let $O$ be the circumcenter of $(BAD)$ and let $K = OC \cap DF$. We have $OD \parallel FC$ since $\angle ODA = 30^{\circ} = \angle DAF$, and we can compute $OD = FC = 1$. So $ODCF$ is a parallelogram and $K$ is the midpoint of $OC$. Then we can compute that $KD = KF = KE = \tfrac{1}{2}$, implying $DE \perp EF$.
05.12.2016 20:35
Just to clarify, one gets the value of KE by noting that BOC and EKC are similar with a scale factor of two. (that comes from the parallelogram). So KE= $\frac{1}{2} $ and the conclusion follows.
21.03.2017 17:22
I was doing this problem in EGMO, I came up with a solution that I can't tell if it's bogus or not. Using the first hint from EGMO, through simple computation we see that $DF=FC$ and $\angle DCF=\angle FDC=30^{\circ}$. Let $DC$ intersect the circumcircle $(BDA)$ again at $G$. Also, let the circle with diameter $DF$ intersect $AC$ again at $H$. Let $F'$ be the reflection of $F$ over $HD$. We see that $F'$ lies on $(BDAG)$ and $GAF'$ is a 30, 60, 90 triangle with right angle at $A$. Thus there is a homothety centered at $C$ that takes $\triangle DHF\rightarrow\triangle GAF'$. This homothety also maps the circumcircles of these two triangles in a 1 to 2 ratio (we can easily compute the ratio of the circumradii with LOS). Since $H$ is the midpoint of $AC$, then the intersection of $(DHF)$ with $BC$ closer to $B$ is the midpoint of $BC$. So point $E$ lies on $(DHF)$, and the perpendicular condition is prove.
21.06.2017 01:04
Barycentric Coordinates :D Here, $(x,y,z)$ is a homogenized coordinate while $(x:y:z)$ is not. Let $Q$ be the point on $DC$ for which $\triangle QDA$ is an equilateral triangle. We will use $\triangle QDA$ as our reference triangle instead of $\triangle ABC$, with $Q = (1,0,0)$, $D = (0,1,0)$ and $A = (0,0,1)$. Then it is easy to see $C = (-1,2,0)$, since $D$ is the midpoint of $QC$. The condition $\angle B = 60^\circ$ implies that $B$ lies on the circumcircle $(QDA)$. Thus, if $B=(u,v,w)$, then $vw+uw+uv=0$. Now, we can calculate the points $E$ and $F$, giving \begin{align*} E &= \tfrac{1}{2}B + \tfrac12 C = \left(\tfrac{u-1}{2},\tfrac{v+2}{2},\tfrac{w}{2}\right),\\ F &= \tfrac{1}{3}A + \tfrac23 C = \left(-\tfrac{2}{3},\tfrac43,\tfrac13\right). \end{align*}Now, we can calculate the displacement vectors $\overrightarrow{DE}$ and $\overrightarrow{FE}$. We get \begin{align*} \overrightarrow{DE} &= E - D = \left(\tfrac{3u+1}{6},\tfrac{3v-2}{6},\tfrac{3w-2}{6}\right)\\ &= (3u+1:3v-2:3w-2), \\ \overrightarrow{FE} &= E - F = \left(\tfrac{u-1}{2},\tfrac{v}{2},\tfrac{w}{2}\right)\\ &= (u-1:v:w). \end{align*}Finally, it remains to show that the vectors are perpendicular, which happens iff \[ \big((3v-2)w+(3w-2)v\big)+\big((3u+1)w+(3w-2)(u-1)\big)+\big((3u+1)v+(3v-2)(u-1)\big)=0 \]by the perpendicularity criterion. But the LHS is equal to \[ 6(vw+uw+uv)-4(u+v+w)+4 = 0-4+4=0, \]and we are done.
Attachments:

28.07.2018 18:59
Note that $D$ lies on the perpendicular bisector of $AC$. Let $M$ be the midpoint of $AC$, $DM\perp AC$. It suffices to prove that $\odot(DMF)$ intersects $BC$ at $E$. Consider the homothety at $C$ with scale $2$. $M \to A$, $E \to B$, $D\to D'$, $F \to F'$ and $\odot(DMF)\to\odot(AF'D')$. It now suffices to prove that $D$ lies on on $\odot(AF'D')$. Note that $AF' : F'C = 1 : 2$ and $CD = CM \sec 30^{\circ} = \tfrac{2}{\sqrt3} CM = \tfrac1{\sqrt3} AC$. We claim that $D$ lies on $\odot(AF'D')$. Since \[CD\cdot CD' = 2CD^2 = \tfrac{2}{3} AC^2 = AC\cdot AF'\]Now, note that $AD'D$ is an equilateral triangle, since, $\angle D'AD = \angle D'AF - \angle DAF' = 60^{\circ}$ and $\angle ADD' = \angle DAC + \angle DCA = 60^{\circ}$. Now, $B$ lies on $\odot(AF'D')$ since $\angle ABD = 60^{\circ} = \angle DD'A$. Sidenote: If we let $CD \cap AB = K$. Then $\odot(BDK)$ and $\odot(DEF)$ are tangent to each other and $AD$ is their common tangent.
30.07.2018 11:54
If $K$ was the circumcenter of $\triangle ABD, AK, DK$ are tangent to the $\odot(ADC)$, thus $CK$ is symmedian of $\triangle ADC$ and subsequently median of $\triangle CFD$, thus, if $L$ was the midpoint of $DF$, then $LE=\frac{BK}2$ (actually $CFKD$ is a parallelogram, thus $KL=CL$). But from $AD=DC$ and $\triangle AKD\cong\triangle CFD$ we get $BK=KD=DF$, hence $LE=DL=LF$ and $\triangle DEF$ is $E-$right-angled. Best regards, sunken rock
09.03.2019 20:54
Let $M$ be the midpoint of $AC$ and let $F'$ be the point on $\overline{AC}$ such that $2AF'=F'C$. Since $\angle{ADM}=\angle{ABD}=60^{\circ}$, $MD$ is tangent to $(ABD)$. Since $MD^2=MF'\cdot MA=\frac{AC^2}{12}$, quadrilateral $ABDF'$ is cyclic and $\angle{ABF'}=\angle{ADF'}=30^{\circ}$. Taking a homothety at $C$ by scale factor $\frac{1}{2}$ , we find that $\angle{ABF'}=\angle{MEF}=\angle{MDF}=30^{\circ}$. Thus, quadrilateral $MDEF$ is cyclic and $\angle{DEF}=180^{\circ}-\angle{DMF}=90^{\circ}$, as desired.
09.03.2019 22:40
Let $M_B$ be the midpoint of $AC$ and let $X$ be the foot of perpendicular from $F$ to $DC$, then, $$\frac{M_BF}{FC}=\frac{M_BC-FC}{FC}=\frac{\tfrac{1}{2}AC-\tfrac{1}{3}AC}{\tfrac{1}{3}AC}=\frac{1}{2}=\sin 30^{\circ}=\frac{M_BD}{DC}$$Hence, $\angle M_BDF=\angle FDC=\angle FCD=30^{\circ}$, Let $F'$ be the reflection of $F$ in $M_B$, then, $DF'$ is the angle bisector of $\angle ADM_B$, hence now, $AF'$ $=$ $FF'$ $=$ $FC$ $=$ $FD$ $=$ $F'D$ $\implies$ $\angle ABD$ $=$ $\angle DF'F$ $=$ $60^{\circ}$, hence, $ABDF'$ is cyclic $\implies$ $\angle ABF'$ $=$ $\angle F'BD$ $=$ $30^{\circ}$, since, $DF$ $=$ $FC$ $\implies$ $DX$ $=$ $XC$, hence, $\implies$ $DB||EX$ and $BF' ||EF$, now, $\angle FEX$ $=$ $\angle FEC$ $-$ $\angle XEC$ $=$ $\angle F'BC$ $-$ $\angle DBC$ $=$ $\angle F'BD$ $=$ $30^{\circ}$, therefore, $\angle FDC$ $=$ $\angle FEX$ $=$ $30^{\circ}$ $\implies$ $FDEX$ is cyclic, $\implies$ $\angle FXD$ $=$ $\angle FED$ $=$ $90^{\circ}$
30.05.2020 06:05
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(30cm); 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 = -2.3, xmax = 20, ymin = -7, ymax = 3; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((9.48,2.12)--(19.12,-6.12), linewidth(1) + dbwrru); draw((19.12,-6.12)--(2.64,-6.28), linewidth(1) + wrwrwr); draw((2.64,-6.28)--(9.48,2.12), linewidth(1) + wrwrwr); draw((xmin, -0.5644774419752964*xmin + 4.672808690567666)--(xmax, -0.5644774419752964*xmax + 4.672808690567666), linewidth(0.00001) + white + dotted); /* line */ draw((2.64,-6.28)--(10.833811978464832,-1.4426337818774861), linewidth(1) + wrwrwr); draw((10.833811978464832,-1.4426337818774861)--(9.48,2.12), linewidth(1) + wrwrwr); draw((10.833811978464832,-1.4426337818774861)--(19.12,-6.12), linewidth(1) + wrwrwr); draw((xmin, -1.693858105629436*xmin + 7.55004592578608)--(xmax, -1.693858105629436*xmax + 7.55004592578608), linewidth(0.00001) + white); /* line */ draw((xmin, 0.3800031272794423*xmin-3.5209729303303585)--(xmax, 0.3800031272794423*xmax-3.5209729303303585), linewidth(0.00001) + white); /* line */ draw((5.3383604845092725,-1.492379251671836)--(2.64,-6.28), linewidth(1) + wrwrwr); draw((5.3383604845092725,-1.492379251671836)--(10.833811978464832,-1.4426337818774861), linewidth(1) + wrwrwr); draw((5.3383604845092725,-1.492379251671836)--(9.48,2.12), linewidth(1) + dbwrru); draw((xmin, -0.5644774419752964*xmin + 1.5210048189659622)--(xmax, -0.5644774419752964*xmax + 1.5210048189659622), linewidth(1) + white); /* line */ draw((5.3383604845092725,-1.492379251671836)--(13.630832927947854,-6.173292884194681), linewidth(1) + wrwrwr); draw((5.3383604845092725,-1.492379251671836)--(19.12,-6.12), linewidth(1) + dbwrru); draw((10.833811978464832,-1.4426337818774861)--(13.630832927947854,-6.173292884194681), linewidth(1) + dtsfsf); draw((10.833811978464832,-1.4426337818774861)--(14.3,-2), linewidth(1) + dtsfsf); draw((14.3,-2)--(13.630832927947854,-6.173292884194681), linewidth(1) + dtsfsf); /* dots and labels */ dot((9.48,2.12),dotstyle); label("B", (9.56,2.32), NE * labelscalefactor); dot((2.64,-6.28),dotstyle); label("A", (2.72,-6.08), NE * labelscalefactor); dot((19.12,-6.12),dotstyle); label("$C$", (19.2,-5.92), NE * labelscalefactor); dot((10.833811978464832,-1.4426337818774861),linewidth(4pt) + dotstyle); label("$D$", (10.92,-1.28), NE * labelscalefactor); dot((5.3383604845092725,-1.492379251671836),linewidth(4pt) + dotstyle); label("$O$", (5.42,-1.34), NE * labelscalefactor); dot((13.630832927947854,-6.173292884194681),linewidth(4pt) + dotstyle); label("$F$", (13.72,-6.02), NE * labelscalefactor); dot((14.3,-2),linewidth(4pt) + dotstyle); label("$E$", (14.38,-1.84), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I totally did not read all of Evan’s hints [I read all of them - every single one] Let $\angle OBD = \angle ODB = x$ and $\angle OAD = \angle ODA = y$. We know that: $$\angle OBD + \angle BDA + \angle DAO + \angle AOB = x + x + y + y + \angle AOB = 360 \Longleftrightarrow \angle AOB = 360 -2x - 2y$$This means that $\angle OBA = \angle OAB = x + y - 90$. Note that: $$\angle OBA + \angle ABD = x + y - 90 + 60 = x + y - 30 = x$$Hence, $y = 30$. Since $\angle ODA = 30 $ and $\angle ADC = 120$ with $\angle DCA = 30$, $OD || FC$. Now, assume WLOG $AC =6$ (if this is not the case, we can always scale the diagram). Then, $AF = 4, FC = 2, AD = 2 \sqrt{3}$. Given that $\angle ODA = \angle OAD = 30$, we can deduce that $OB = OD = OA = 2$. Hence, $OD = FC$, so $ODCF$ is a parallelogram. By Stewart’s theorem: $$DF^2 (CA) + CF \cdot FA \cdot CA = DC^2 \cdot FA + DA^2 \cdot CF \Longleftrightarrow DF = 2$$Note that the intersection of $OC$ and $DF$ is the midpoint of $DF$ ($ODCF$ is a parallelogram). The homothety that takes $E$ to $B$ (from $C$) also takes the midpoint of $DF$ to $O$, so the distance from the midpoint of $DF$ to $E$ is $\frac{BO}{2} = 1$. It follows that $DEF$ is a right triangle.
29.04.2021 06:01
Here's a purely synthetic solution. Let the reflection of $C$ over $D$ be $C'$, the midpoint of $AC$ be $M$ and the point at which the circumcircle of $ABD$ intersects $AC$ be $G$. It's immediate that $\angle DGA=180^\circ-\angle DBA=120^\circ$; since $G$ lies on $AC$, $\angle GAD=30^\circ=\angle GDA$ and we have that $AG=\frac{AD}{\sqrt{3}}=\frac{AC}{3}$, so $GC=\frac{2AC}{3}$. Since $\angle AC'D=\angle DAC+\angle DCA=60^\circ=\angle ABD$, $C'$ lies on the circumcircle of $ABD$. Now, consider the homothety sending the circumcircle of $ABG$ to the circumcircle of $MEF$ centered at $C$. It has scale factor $\frac{1}{2}$, so it maps $C'$ to $D$, and hence $DEMF$ is cyclic. This gives $\angle DEF=\angle DMF=90^\circ$ as desired.
29.04.2021 06:49
posted here
11.06.2021 05:58
First I misread thinking $F$ was not on segment $BC$. Then I misread that $D$ is outside $\triangle ABC$. What am I doing? [asy][asy] size(250); pair A = origin, C = (3, 0), D = (3/2, sqrt(3)/2); pair X = extension(C, D, (0, 0), (0, 50)); pair B = circumcenter(A, D, X)+abs(circumcenter(A, D, X)-A)*dir(87); draw(A--C--D--B--cycle); pair EE = (B+C)/2; pair F = C*2/3; draw(B--C); draw(D--EE--F--cycle, dashed); pair M = (D+F)/2; draw(EE--M); draw(EE--circumcenter(A, B, D)--F); draw(circumcenter(A, B, D)--B); draw(circumcenter(A, B, D)--C, dashed); draw(circumcenter(A, B, D)--D--A); dot("$O$", circumcenter(A, B, D), SW); dot("$A$", A, SW); dot("$C$", C, SE); dot("$B$", B, N); dot("$F$", F, S); dot("$M$", M, SW); dot("$E$", EE, NE); dot(D); label("$D$", D+(0, 0.05), N); [/asy][/asy] Let $M$ be the midpoint of $DF$. It suffices to show that $EM = \frac 12 DF$. WLOG let $AC=3$. From the Law of Cosines, we obtain $$DF = \sqrt{1+3-\sqrt 3 \cdot \frac{\sqrt 3}2 \cdot 2} = 1,$$so it suffices to show that $EM = \frac 12$. First, verify that the circumradius of $\triangle ABD$ is 1 by the Extended Law of Sines, so compute $OB=1$. Furthermore, since $\angle ODA = 30^\circ = \angle DAC$ and $OD=CF=1$, $ODCF$ is a parallelogram. It follows that there is a homothety at $M$ taking $FC$ to $GD$ with ratio $-1$, and hence $M$ is the midpoint of $CG$. Therefore, there is a homothety at $C$ with ratio $\frac 12$ taking $BG$ to $ME$. It follows that $ME = \frac 12$ for all choices of $B$, and $DE \perp EF$ as desired. $\square$
30.06.2021 02:34
Let $F'$ be the reflection of $F$ over $M$ and $C'$ be the reflection of $C$ over $D$ and $M$ be the midpoint of $AC$. It is trivial that $\angle DMA=90$. Now if $DM=x$ then $AM=x\sqrt{3}$ and $F'M=\frac{x\sqrt{3}}{2}$ and since $DM \perp AM$, we have $DF'M$ is a 30-60-90 triangle. This quickly gives that $AF'DB$ is cyclic and that $F'A=F'D$. In addition we get $\angle F'DC=90$. This means $\angle F'CD=\angle F'C'D=30$ but since $\angle F'DA=30$, we have that $C'$ lies on $(ADB)$. Finally, a homothety of ratio $0.5$ from $C$ sends $A,F',B,C'$ to $M,F,E,D$ respectively, so they are all cyclic. And thus \[\angle DEF=180-\angle DMF=90\]so we are done.
30.06.2021 03:54
Consider point $X$ outside triangle $DAC$ such that $XA = XD$ and $\angle AXD = 120^{\circ}$. Consider the circle centered at $X$ with radius $PA$, which we let equal 1. Denote this circle as $\omega$. We then consider the configuration on the complex plane, letting $X$ be $0$ (or the center/origin), making $\omega$ the unit circle. Since $\angle DBA = 60^{\circ}$, we know that $B$ lies on $\omega$. Some raw computation yields that $d = 1$, $f = \frac32 + \frac{\sqrt{3}}{2}i$, $c = \frac52 + \frac{\sqrt{3}}{2}i$, and $e = \frac{b+c}{2}$. We will keep these as reference, as these will be useful later on. In order to show that $DE \perp EF$, it suffices to show that $V = \frac{d-e}{e-f} + \overline{\left(\frac{d-e}{e-f}\right)} = 0$. We substitute the values above that were kept as reference to obtain that$$V = \frac{2-b-c}{b+c-2f} + \frac{2-\overline{b} - \overline{c}}{\overline{b} + \overline{c} - 2\overline{f}}$$We know that $2-c = e^{\frac{2\pi i}{3}}$, $2 - \overline{c} = e^{\frac{4\pi i}{3}}$, $c-2f = e^{\frac{2\pi i}{3}}$, and $\overline{c} - 2\overline{f} = e^{\frac{4\pi i}{3}}$ from substitution. Therefore, we get the following:$$V = \frac{e^{\frac{2\pi i}{3}} - b}{e^{\frac{2\pi i}{3}} + b} + \frac{e^{\frac{4\pi i}{3}} - \overline{b}}{e^{\frac{4\pi i}{3}} + \overline{b}}$$Combining denominators and simplifying, the numerator becomes$$(e^{\frac{2\pi i}{3}} - b)(e^{\frac{4\pi i}{3}} + \overline{b}) + (e^{\frac{2\pi i}{3}} + b)(e^{\frac{4\pi i}{3}} - \overline{b})$$It turns out that this expression is nice AF and everything CANCELS OUT, so therefore $V = 0$, as desired.
30.06.2021 05:38
A fairly motivated synthetic solution:
30.06.2021 10:07
$G$ is the midpoint of $AF$. $H$ is on $BG$ or the extension of $BG$, and $CH \perp BH$. $\angle ABD = 60 ^{\circ} = \angle DGC $ $ \Longrightarrow$ $AGDB$ are concyclic $ \Longrightarrow$ $\angle HBD= \angle DAG =30 ^{\circ}$ $\angle GDC = 90 ^{\circ} = \angle GHC $ $ \Longrightarrow$ $GHCD$ are concyclic $ \Longrightarrow$ $\angle DHC= \angle DGC =60 ^{\circ}$ The extension of $BD$ crosses $HC$ at $M$. $ \Longrightarrow$ $BD=HD=DM$ $ \Longrightarrow$ $DE//HC$ $ \Longrightarrow$ $DE \perp BH$ $ \Longrightarrow$ $DE \perp EF$
06.09.2021 21:42
11.12.2021 16:23
31.08.2023 00:48
Very nice problem! We'll show that E lies on the circle with diameter DF=$\omega$. Note that the midpoint of AC=M lies on $\omega$ since ADC is isosceles, $N=CD\cap\omega$ is the midpoint of CD since CFD is isosceles (proof: DMC is 30-60-90, MF/FC=(1/6)/(1/3)=1/2=MD/DC implies DF bisects angle MDC which is 60 deg), and F is the midpoint of $AC\cap(ABD)=P$ with C, since APD=120 means DPF=60, whence FPD is equilateral implies FP=FD=FC; in particular, (ADP) goes to (MNF), with A going to E by homothety at C; in particular, since B lies on (ADP), E lies on (DNFM), as desired. $\blacksquare$
24.12.2023 04:15
Let $X = CD \cap (ABD)$ and $Y = CA \cap (ABD)$. Then $\triangle AXD$ is equilateral. $\triangle AYD$ is a 30-30-120 triangle, so $\angle DAY = 30$ and $F$ is the midpoint of $YC$. Thus there exists a homothety of scale factor 2 mapping $\triangle DEF \rightarrow \triangle XBY$, so \[\angle DEF = \angle XBY = 180 - \angle YAX = 180-60-30 = 90. \quad \blacksquare\]
02.01.2024 05:25
Let $D=(2, 0)$, $A=(-1, -\sqrt3)$, $C=(5, -\sqrt3)$, $F=(3, -\sqrt3)$, and $B=(2a, 2b)$ where $a^2+b^2=1$. We see that $E=(a+\frac52, b-\frac{\sqrt3}{2})$ so we must have that $\frac{(a-\frac12)+(b+\frac{\sqrt3}2)i}{(a+\frac12)+(b-\frac{\sqrt3}2)i}\in i\mathbb{R}$. Since $a^2+b^2=1$, $\text{Re}(\frac{(a-\frac12)+(b+\frac{\sqrt3}2)i}{(a+\frac12)+(b-\frac{\sqrt3}2)i})=\frac{(a^2-\frac14)+(b^2-\frac34)}{(a+\frac12)^2+(b-\frac{\sqrt3}2)^2}=0$, as desired.
21.08.2024 21:51
lol what Note that $\triangle ADC\sim\triangle DFC$. Therefore, the circle with diameter $DF$ passes through the midpoints of $AC$ and $DC$. Thus, taking a homothety with scale factor $2$ from $C$ will take this circle to the circle with $A$, $D$, and the reflection of $C$ over $D$. Therefore, it suffices to show that $B$ is on this circle as well. But this is just because $\angle ABD=60^{\circ}$. $\blacksquare$