Denote the points $ X \in AE \cap BD$, $ Y \in AE \cap BC$, $ Z \in AE \cap DF$ and $ T \in DF \cap BC$. In $ \triangle{AEC}$, we observe that the lines $ AF$, $ DE$ and $ BC$ are concurrent, if and only if, the division $ (AYEZ)$ is harmonic. Since the quadrilateral $ XYTD$ is cyclic, $ \tan{XYB}= \tan{XDZ}$, which is equivalent to \[ \frac{XB}{XY}=\frac{XZ}{XD}\Longleftrightarrow XB \cdot XD = XY \cdot XZ. \] Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2}=XB \cdot XD$, so $ XA^{2}=XY \cdot XZ$, which by using $ XA=XF$, it is equivalent with $ \frac{YA}{YE}=\frac{ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic.

We denote as $K,$ $L,$ the intersection points of $BC,$ from $DF,$ $AE$ respectively and also the points $X\equiv AC\cap BE,$ $Y\equiv AB\cap CE,$ $Z\equiv AE\cap XY.$ From the complete quadrilateral $ABECXY,$ we conclude that the points $L,$ $Z,$ are harmonic conjugates, with respect to the points $A,$ $E.$ Because of $AD\perp AB,$ $DE\perp BE,$ $DK\perp KB,$ we conclude that the pentagon $ABEKD$ is cyclic ( taken as diameter the segment $BD$ ) and so, we have that the segment line $KB,$ is the angle bisector of the angle $\angle AKE,$ because of $AB = BE.$ So, because of $DK\perp KB,$ we conclude that the segment line $DF,$ passes through the point $Z,$ as the external angle bisector of the triangle $\bigtriangleup KAE,$ through vertex $K.$ Because of now, the points $X,$ $Y,$ $Z$ are collinear, based on the Desarques’s theorem, we conclude that the triangles $\bigtriangleup BEA,$ $\bigtriangleup CDF,$ are perspective $($ $X\equiv BE%Error. "capCD" is a bad command. ,$ $Y\equiv BA\cap CF,$ $Z\equiv AE\cap DF$ $).$ Hence, the segment lines $AF,$ $DE,$ $BC$ are concurrent at one point and the proof is completed. Kostas Vittas.

Here is another solution to this nice problem: Let $DF\cap BC= O$, $AE\cap BD=M$, $AE\cap BC=N$ and assume $AD\geq DC$. Let $\measuredangle{CBD}=x$, $\measuredangle{DBA}=y$, and $\measuredangle{BCA}=c$. It`s clear that $x+y+c=90^{\circ}$ Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$. Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$, thus $\measuredangle{OAE}=y-x$. In isosceles triangle $ABE$, we have $\measuredangle{BAE}=c+x$, therefore $\measuredangle{ABE}=2y$. But $\measuredangle{ABN}=x+y$, therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$. This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic. This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$. In triangle $ABE$, $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$ In triangle $EDC$, $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$ By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$, therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED.

pohoatza wrote: Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2} = XB \cdot XD$, so $ XA^{2} = XY \cdot XZ$, which by using $ XA = XF$, it is equivalent with $ \frac {YA}{YE} = \frac {ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic. Please could you tell how exactly using $ XA=XF$ we get $ \frac {YA}{YE} = \frac {ZA}{ZE}?$

Remark (a short commentary). It is a very nice application of the following remarkable harmonical division (see the nice Pohoatza's proof !) : Quote: Lemma. Let $ w(O)$ , $ w_o$ be two secant circles in $ \{A,B\}$ so that $ O\in w_o$ . For $ P\in (AB)$ (segment !) denote $ \{M,N,R\}$ such that $ \{M,N\}\subset w$ , $ R\in w_o$ and $ N$ separates $ P$ , $ R$ . Then the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically. Particular case. Let $ ABC$ be a triangle with the orthocenter $ H$ . Denote $ D\in BC\cap AH$ and the intersections $ N$ , $ S$ between the line $ AH$ and the circle with the diameter $ [BC]$ . Then the division $ \{\ A\ ,\ H\ ,\ N\ ,\ S\ \}$ is harmonically. Proof. $ PO\cdot PR = PA\cdot PB = PM\cdot PN\ \implies\ PO\cdot PR = PM\cdot PN\ \Longleftrightarrow$ the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically. The following interesting problem is a nice consequence of the harmonical division which was mentioned in the above lemma : Quote: Let $ w(O)$ , $ w_o$ be two secant circles so that $ O\in w_o$ . Consider $ A\in w\cap w_o$ . For $ P\in (AB)$ (segment !) denote $ \{M,N,R\}$ such that $ \{M,N\}\subset w$ , $ R\in w_o$ and $ N$ separates $ P$ , $ R$ . For $ C\in AB$ (line !) denote $ \left\|\begin{array}{c} X\in MC\cap RA \\ \ Y\in NC\cap RA\end{array}\right\|$ .Then $ MY\cap NX\cap CP\ne\emptyset$ . Proof. From the upper lemma obtain that the division $ \{\ M\ ,\ N\ ;\ P\ ,\ R\ \}$ is harmonically, i.e. $ \frac {\overline {PM}}{\overline {PN}} = - \frac {\overline {RM}}{\overline{RN}}$ . Apply the Menelaus' theorem to the transversal $ \overline {RXY}$ and the triangle $ CMN\ : \ \frac {\overline{RM}}{\overline{RN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} = + 1$ $ \implies$ $ \frac {\overline{PM}}{\overline{PN}}\cdot \frac {\overline{YN}}{\overline{YC}}\cdot \frac {\overline{XC}}{\overline{XM}} = - 1$ , i.e. $ MY\cap NX\cap CP\ne\emptyset$ .

Sergey wrote: pohoatza wrote: Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2} = XB \cdot XD$, so $ XA^{2} = XY \cdot XZ$, which by using $ XA = XF$, it is equivalent with $ \frac {YA}{YE} = \frac {ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic. Please could you tell how exactly using $ XA = XF$ we get $ \frac {YA}{YE} = \frac {ZA}{ZE}?$ There is a typo. He means $ XA = XE,$ instead of $ XA = XF.$ Kostas Vittas.

Hello! $ \left\{M\right\}=BE\cap DF$, $ \left\{N\right\}=BC\cap DF$, $ \left\{H\right\}=DE\cap BN$, $ \left\{S\right\}=MH\cap BD$, $ \left\{T\right\}=MH\cap AC$. $ DE\bot MB$ and $ BN\bot MD\Rightarrow$ the point $ H$ is the orthocenter of the triangle $ \Delta MBD\Rightarrow$ $ \Rightarrow MS\bot BD$ and $ \angle MSN=\angle MSE$ (1) $ \Delta ABS\equiv\Delta EBS\Rightarrow\angle ASB=\angle ESB\Rightarrow\angle AST=\angle MSE$ (2) (1), (2) $ \Rightarrow\angle MSN=\angle AST\Rightarrow$ the points $ A,S,N$ are collinear. Let's consider the triangles $ \Delta SDA$ and $ \Delta MEF$. $ \left\{B\right\}=SD\cap ME$, $ \left\{C\right\}=DA\cap EF$, $ \left\{N\right\}=AS\cap FM$ and the points $ B,C,N$ are collinear. Based on the Desargues's theorem, we conclude that the triangles $ \Delta SDA$ and $ \Delta MEF$ are perspective (the lines $ SM,DE$ and $ AF$ are concurrent at the point $ H$). Hence, the lines $ AF,DE$ and $ BC$ are concurrent at the point $ H$. Best regards, Petrisor Neagoe

PhilAndrew wrote: Let $ ABC$ be a right triangle with $ A = 90^{\circ}$ and $ D \in (AC)$. Denote by $ E$ the reflection of $ A$ in the line $ BD$ and $ F$ the intersection point of $ CE$ with the perpendicular in $ D$ to $ BC$. Prove that $ AF, DE$ and $ BC$ are concurrent. Here is another approach: Let $ T\equiv DF \cap AE$, $ S\equiv AE\cap BC$, $ K\equiv AE\cap BD$, $ H\equiv DT\cap BC$. In order to prove $ AF,DE,BC$ are concurrent, we need to prove $ (A,E,S,T) = - 1$. Define $ (BD)$ is the circle with diameter $ BD$, $ (DS)$ is the circle with diameter $ DS$. It is easy to notice that $ \{A,B,D,E,H\}\in (BD)$ and $ \{K,D,H,S\}\in (DS)\Longrightarrow DH$ is the radical axis wrt $ (BD),(DS)$. Thus, $ \mathcal {P}_{T/(BD)} = \mathcal {P}_{T/(DS)}$ $ \Longrightarrow \overline{TS}.\overline {TK} = \overline {TE}.\overline {TA}$, but $ K$ is the midpoint of $ AE$, hence $ (AEST) = - 1$, which implies to the result of the problem. Our proof is completed $ \blacksquare$.

andyciup wrote: Here is another solution to this nice problem: Let $DF\cap BC= O$, $AE\cap BD=M$, $AE\cap BC=N$ and assume $AD\geq DC$. Let $\measuredangle{CBD}=x$, $\measuredangle{DBA}=y$, and $\measuredangle{BCA}=c$. It`s clear that $x+y+c=90^{\circ}$ Because triangle $BAM$ is right-angled, we get $\measuredangle{BAM}=c+x$. Because the quadrilateral $ABDO$ is cyclic, we have $\measuredangle{OAD}=\measuredangle{OBD}=x$, thus $\measuredangle{OAE}=y-x$. In isosceles triangle $ABE$, we have $\measuredangle{BAE}=c+x$, therefore $\measuredangle{ABE}=2y$. But $\measuredangle{ABN}=x+y$, therefore $\measuredangle{EBO}=y-x=\measuredangle{OAE}$. This means that quadrilaterel $OABE$ is cyclic, and thus the pentagon $ADOEB$ is cyclic. This means that $\measuredangle{EDB}=\measuredangle{EAB}=c+x\Longrightarrow \measuredangle{ADE}=2(c+x)\Longrightarrow \measuredangle{EDF}=y-x$. In triangle $ABE$, $\ds N\in (AE)\Longrightarrow \frac{AN}{NE}=\frac{AB}{BE}\cdot\frac{\sin \measuredangle{ABN}}{\sin\measuredangle{EBN}}\Longrightarrow \frac{AN}{NE}=\frac{\sin(x+y)}{\sin(y-x)}$ In triangle $EDC$, $\ds F\in (EC)\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin\measuredangle{EDF}}{\sin\measuredangle{FDC}}\Longrightarrow \frac{EF}{FC}=\frac{ED}{DC}\cdot\frac{\sin(y-x)}{\sin(x+y)}$ By multiplying the last two relations we obtain $\ds \frac{AN}{NE}\cdot\frac{EF}{FC}\cdot\frac{CD}{DA}=\frac{ED}{DA}=1$, therefore, by the converse of Ceva`s theorem, the lines $AF, CN, ED$ are concurent, QED. Darn it this was exactly the same proof I just rediscovered a few minutes ago, which I originally wrote down a while ago but threw away the paper.

PhilAndrew wrote: Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent. The inspiration for this solution comes from Cosmin (pohoatza) in https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCAQFjAA&url=http%3A%2F%2Fdiendantoanhoc.net%2Fforum%2Findex.php%3Fapp%3Dcore%26module%3Dattach%26section%3Dattach%26attach_id%3D16951&ei=dokOVOH4E8S0yASt5YLgBQ&usg=AFQjCNHsxpkOva9b0YPm1Io5LkgocHpOCg&bvm=bv.74649129,d.aWw (Harmonic Divisions and its Applications).

PhilAndrew wrote: Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent. My solution. Let $X$,$Y$, and $Z$ be the intersections of the line $AE$ with the lines $BD$,$BC$, and $DF$, respectively, and let $W$ be the intersection of the lines $DF$ and $BC.$ Because $E$ is the reflection of $A$ in the line $BD$ we have, $\angle DEB=\angle BAD=\angle DWB = 90^\circ$, which means that the pentagon $ADWEB$ is Cyclic. Thus, we have $$\angle YWE=\angle BWE= \angle BAE=\angle BAX=90^\circ - \angle XAD= \angle ADB=\angle AWB=\angle AWY$$and we know that the lines $YW$ and $WZ$ are perpendicular. Therefore, the cross ratio $(A,E,Y,Z)$ is harmonic bundle, which means that the lines $CB$, $DE$, and $AF$ are concurrent.

Sergey wrote: pohoatza wrote: Due to the similarity of the triangles $ \triangle{XAB}$ and $ \triangle{XDA}$, we have that $ XA^{2} = XB \cdot XD$, so $ XA^{2} = XY \cdot XZ$, which by using $ XA = XF$, it is equivalent with $ \frac {YA}{YE} = \frac {ZA}{ZE}$, i.e. the division $ (AYEZ)$ is harmonic. Please could you tell how exactly using $ XA=XF$ we get $ \frac {YA}{YE} = \frac {ZA}{ZE}?$ it is newton theorem

Let $DF \cap BC = K $ It is clear that $ADBK $ and $ADBE $ are cyclic which forces $AKEB $ also to be cyclic. Now ,since $\angle EAB = \angle AEB ,BK $ bisects $\angle AKE $. Now,by blanchet's theorem in $\triangle CDF $,$AF,DE,BC $ concur. $\blacksquare$

Ig there was a need for a different solution. So Bumped. PhilAndrew wrote: Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent. Let $DF\cap BC=T$. So, $\{A,B,E,T,D\}$ all lie on a circle. Now we will use Ratio Lemma and Ceva's Theorem. $\frac{AK}{KE}=\frac{AB\sin\angle BAC}{BE\sin\angle EBC}$ and $\frac{EF}{FC}=\frac{ED\sin\angle EDF}{CD\sin\angle FDC}$. Now $\angle EDF=\angle EBC$ and $\angle ABC=\angle FDC$. So now applying Ceva's Theorem we get that $$\frac{AD}{DC}\cdot\frac{CF}{FE}\cdot\frac{EK}{KA}=\frac{AD}{DC}\cdot\frac{DC\sin\angle FDC}{ED\sin\angle EDF}\cdot\frac{BE\sin\angle EBC}{AB\sin\angle ABC}=\frac{AD}{DC}\cdot\frac{DC}{ED}\cdot{BE}{AB}=1\implies AF,DE,BC\text{ are concurrent.}\blacksquare$$

Let $P=AF\cap BC$, $Q=DF\cap BC$ and $H=AE\cap BC$. See that $Q\in (ABDE)$ and therefore $$(Q,H,C,P)\overset{A}{=}(G,E,C,F)\overset{Q}{=}(A,E,B,D)=-1.$$Furthermore, if $P'=DE\cap BC$ and $I=DH\cap (ABD)$ $$(Q,H,C,P')\overset{D}{=}(Q,I,A,E)\overset{H}{=}(B,D,E,A)=-1,$$and thus $P=P'$, as desired. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.718181818181812cm); 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 = -12.187272727272727, xmax = 15.249090909090897, ymin = -6.133636363636364, ymax = 6.957272727272726; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw((-1.97908720787453,1.204050950799106)--(-1.6605867761802946,0.9865243878485139)--(-1.4430602132297026,1.3050248195427494)--(-1.761560644923938,1.5225513824933414)--cycle, linewidth(0.6)); draw((-6.394305392080065,-2.59)--(-6.394305392080065,-2.204305392080065)--(-6.78,-2.204305392080065)--(-6.78,-2.59)--cycle, linewidth(0.6)); draw(arc((-6.78,4.95),1.,-90.,-73.66615466770288)--(-6.78,4.95)--cycle, linewidth(0.6)); draw(arc((-6.78,4.95),1.,-73.6661546677029,-34.33194359599434)--(-6.78,4.95)--cycle, linewidth(0.6)); draw(arc((4.26,-2.59),1.4545454545454537,530.2897748075213,540.)--(4.26,-2.59)--cycle, linewidth(0.6)); draw(arc((4.26,-2.59),1.4545454545454537,505.66805640400565,530.2897748075213)--(4.26,-2.59)--cycle, linewidth(0.6)); /* draw figures */ draw((-6.78,4.95)--(-6.78,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(-4.570314034778992,-2.59), linewidth(0.7) + sexdts); draw((-6.78,-2.59)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-4.570314034778992,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-6.78,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-4.570314034778992,-2.59)--(-1.761560644923938,1.5225513824933414), linewidth(0.7) + sexdts); draw((4.26,-2.59)--(-3.6463196618288167,-1.2370957722321034), linewidth(0.7) + sexdts); /* dots and labels */ dot((-6.78,4.95),dotstyle); label("$B$", (-6.714545454545459,5.139090909090909), NE * labelscalefactor); dot((-6.78,-2.59),dotstyle); label("$A$", (-7.332727272727277,-3.1518181818181823), NE * labelscalefactor); dot((4.26,-2.59),dotstyle); label("$C$", (4.34,-2.4063636363636367), NE * labelscalefactor); dot((-4.570314034778992,-2.59),dotstyle); label("$D$", (-4.423636363636369,-3.17), NE * labelscalefactor); dot((-2.7101665988177412,-1.3972873014084721),dotstyle); label("$E$", (-2.5509090909090975,-1.97), NE * labelscalefactor); dot((-3.6463196618288167,-1.2370957722321034),linewidth(4.pt) + dotstyle); label("$F$", (-3.86,-0.9518181818181822), NE * labelscalefactor); dot((-0.01585517068551292,0.3302851437471739),linewidth(4.pt) + dotstyle); label("$X$", (0.12181818181817432,0.5209090909090904), NE * labelscalefactor); label("$\alpha$", (-6.696363636363641,3.6118181818181814), NE * labelscalefactor); label("$\beta$", (-6.187272727272732,3.757272727272727), NE * labelscalefactor); label("$\theta_1$", (2.067272727272719,-2.5790909090909095), NE * labelscalefactor); label("$\theta_2$", (2.1945454545454464,-1.8609090909090913), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here's a trig-ceva solution: Suppose $\angle ABD = \alpha$, and $\angle DBC = \beta$. Also, let $\angle ECD = \theta_1$, and $\angle ECB = \theta_2$. Now, trig-ceva on $\displaystyle \triangle BDC$ for the point $E$ gives $$\frac{\sin \theta_1}{\sin \theta_2} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \ \ \ \ \ -(1) $$ Let $\angle FBC = x$. Trig-Ceva on $\displaystyle \triangle BDC$ for the point $F$ gives $$\frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin(x)}{\sin(\beta -x)} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta} = 1 \implies \frac{\sin(\beta - x)}{\sin(x)} = \frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin(\alpha+ \beta)}{\cos \beta} $$where we used $(1)$ in the last step. Thus, we get $$\sin \beta \cdot \cot(x) = \cos \beta + \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\alpha+\beta) \cdot \sin (\beta - \alpha)} \ \ \ \ \ -(2) $$ Let $\angle FAC = y$. Now, trig-ceva on $\displaystyle \triangle ABC$ for the point $F$ gives $$\frac{\sin y}{\cos y} = \frac{\sin \theta_1}{\sin \theta_2} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} = \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} \ \ \ \ \ -(3) $$where we used $(1)$ in the last equality. Thus, finally we have by Trig Ceva in $\displaystyle \triangle ABD$ for the lines $AF, DE, BC$, that $AF,DE,BC$ are concurrent if and only if $$\frac{\sin y}{\cos y} \cdot \frac{\sin(\alpha+\beta)}{\sin \beta} \cdot \frac{\cos \alpha}{\sin(2 \alpha)} = 1 \iff \frac{\sin(2 \alpha)}{\cos \alpha} \cdot \frac{\sin \alpha}{\sin(\beta - \alpha)} \cdot \frac{\sin x}{\sin(\alpha + \beta -x)} \cdot \frac{\sin(\alpha+\beta)}{\sin \beta} \cdot \frac{\cos \alpha}{\sin(2 \alpha)} = 1 $$Cancelling out terms, and taking $\sin \beta$ and terms involving $x$ to the other side, we are reduced to showing that $$ \sin \beta \cdot\sin(\alpha + \beta) \cdot \cot x - \sin \beta \cdot \cos(\alpha + \beta) = \frac{\sin \alpha \cdot \sin(\alpha+\beta)}{\sin(\beta - \alpha)}$$Plugging in the value of $\sin \beta \cdot \cot x$ from $(2)$, we get $$LHS = \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\beta - \alpha)} + \sin(\alpha + \beta) \cdot \cos \beta - \sin \beta \cdot \cos(\alpha + \beta) = \frac{2 \cdot \sin ^2 \alpha \cdot \cos \beta}{\sin (\beta - \alpha)} + \sin \alpha = \frac{\sin \alpha \cdot \sin(\alpha+\beta)}{\sin(\beta - \alpha)}.$$Thus, we are done. $\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.230909090909087cm); 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 = -12.18727272727273, xmax = 15.249090909090896, ymin = -6.133636363636364, ymax = 6.957272727272726; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw((-1.97908720787453,1.204050950799106)--(-1.6605867761802946,0.9865243878485139)--(-1.4430602132297026,1.3050248195427494)--(-1.761560644923938,1.5225513824933414)--cycle, linewidth(0.6)); draw((-6.394305392080065,-2.59)--(-6.394305392080065,-2.204305392080065)--(-6.78,-2.204305392080065)--(-6.78,-2.59)--cycle, linewidth(0.6)); /* draw figures */ draw((-6.78,4.95)--(-6.78,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(-4.570314034778992,-2.59), linewidth(0.7) + sexdts); draw((-6.78,-2.59)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-4.570314034778992,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-6.78,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-4.570314034778992,-2.59)--(-1.761560644923938,1.5225513824933414), linewidth(0.7) + sexdts); draw((4.26,-2.59)--(-3.6463196618288167,-1.2370957722321034), linewidth(0.7) + sexdts); /* dots and labels */ dot((-6.78,4.95),dotstyle); label("$B$", (-6.714545454545459,5.139090909090909), NE * labelscalefactor); dot((-6.78,-2.59),dotstyle); label("$A$", (-7.332727272727277,-3.1518181818181823), NE * labelscalefactor); dot((4.26,-2.59),dotstyle); label("$C$", (4.34,-2.4063636363636367), NE * labelscalefactor); dot((-4.570314034778992,-2.59),dotstyle); label("$D$", (-4.423636363636369,-3.17), NE * labelscalefactor); dot((-2.7101665988177412,-1.3972873014084721),dotstyle); label("$E$", (-2.5509090909090975,-1.97), NE * labelscalefactor); dot((-3.6463196618288167,-1.2370957722321034),linewidth(4.pt) + dotstyle); label("$F$", (-3.86,-0.9518181818181822), NE * labelscalefactor); dot((-0.01585517068551292,0.3302851437471739),linewidth(4.pt) + dotstyle); label("$X$", (0.12181818181817432,0.5209090909090904), NE * labelscalefactor); dot((-1.761560644923938,1.5225513824933414),linewidth(4.pt) + dotstyle); label("$G$", (-1.6963636363636432,1.6663636363636358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here's a coordinate solution: Toss the figure in the coordinate plane. Let $G$ be the foot of perpendicular from $D$ to $BC$. We may assume without loss of generality, that $$G = (0,0), \ B = (b,0), \ C = (c,0), \ D = (0,d).$$Now, let $A = (\beta, \alpha)$. We have $$\frac{\alpha}{d} + \frac{\beta}{c} = 1 \ \ \ \ \ \ \ -(1) \ \ \ \ \ \alpha = \frac cd (\beta - b) \ \ \ \ \ \ \ -(2)$$because $A$ lies on $CD$, and because $A$ also lies on the perpendicular from $B$ to $CD$. Solving $(1)$ and $(2)$, we get $$\alpha = \frac{cd(c-b)}{(c^2+d^2)}, \ \ \ \beta = \frac{c(d^2+bc)}{(c^2 + d^2)}.$$ Now, let $E = (e_x, e_y)$. The midpoint of $AE$ lies on $BD$, so $\frac{e_y+\alpha}{2d} + \frac{e_x+\beta}{2b} = 1$. Also, $AE \perp BD$, so $e_y - \alpha = \frac{b}{d} \cdot (e_x - \beta)$. Solving these for $e_x$, we get (note that $d- \alpha = \frac{d(d^2 + bc)}{(c^2 + d^2)}$) $$\left(\frac bd + \frac db \right) e_x = \left(\frac bd - \frac db \right) \beta + 2(d- \alpha) \implies \frac{e_x}{\beta} = \frac{b^2 - d^2}{b^2 + d^2} + \frac{2bd(d - \alpha)}{\beta (b^2 + d^2)} = \frac{b^2 - d^2}{b^2 + d^2} + \frac{2bd^2}{c(b^2 + d^2)} = \frac{\gamma}{c(b^2 + d^2)} = \frac{1}{\kappa},$$where $\gamma = c(b^2 - d^2) + 2bd^2$ and $\kappa = \frac{c(b^2 + d^2)}{\gamma}$. Now, (using that $\frac{\beta}{\alpha} = \frac{d^2+bc}{d(c-b)}$) $$\frac{e_y}{\alpha} = 1 + \frac{\beta}{\alpha} \cdot \frac bd \cdot \left( \frac{\gamma}{c(b^2 + d^2)} - 1 \right) = 1+ \frac{b(d^2+bc)}{d^2(c-b)} \cdot \frac{2d^2(b-c)}{c(b^2 + d^2)} = 1 - \frac{2b(d^2 + bc)}{c(b^2 + d^2)} = - \frac{\gamma}{c(b^2 + d^2)} = - \frac{1}{\kappa}.$$ Now, let $F = (0,f)$ and let $X = (x,0)$ be the intersection of $DE$ with $BC$. Then $$\frac{e_x}{c} + \frac{e_y}{f} = 1 \implies \frac{\alpha}{f} = \left(1 - \frac{e_x}{c} \right) \frac{\alpha}{e_y} = \left( 1 - \frac{\beta}{c \kappa} \right) (-\kappa) = - \kappa + \frac{\beta}{c}.$$Also, $$\frac{e_x}{x} + \frac{e_y}{d} = 1 \implies \frac{\beta}{x} = \left(1 - \frac{e_y}{d} \right) \frac{\beta}{e_x} = \left( 1 + \frac{\alpha}{d \kappa} \right) (\kappa) = \kappa + \frac{\alpha}{d}.$$Thus, it suffices to show that $A$ satisfies the equation of $FX$, i.e. $$\frac{\alpha}{f} + \frac{\beta}{x} = 1 \iff - \kappa + \frac{\beta}{c} + \kappa + \frac{\alpha}{d} = 1$$which is true because of $(1)$. Thus, we are done. $\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.230909090909087cm); 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 = -12.18727272727273, xmax = 15.249090909090896, ymin = -6.133636363636364, ymax = 6.957272727272726; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); draw((-1.97908720787453,1.204050950799106)--(-1.6605867761802946,0.9865243878485139)--(-1.4430602132297026,1.3050248195427494)--(-1.761560644923938,1.5225513824933414)--cycle, linewidth(0.6)); draw((-6.394305392080065,-2.59)--(-6.394305392080065,-2.204305392080065)--(-6.78,-2.204305392080065)--(-6.78,-2.59)--cycle, linewidth(0.6)); /* draw figures */ draw((-6.78,4.95)--(-6.78,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-6.78,4.95)--(-4.570314034778992,-2.59), linewidth(0.7) + sexdts); draw((-6.78,-2.59)--(4.26,-2.59), linewidth(0.7) + rvwvcq); draw((-4.570314034778992,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-6.78,-2.59)--(-0.01585517068551292,0.3302851437471739), linewidth(0.7) + dbwrru); draw((-4.570314034778992,-2.59)--(-1.761560644923938,1.5225513824933414), linewidth(0.7) + sexdts); draw((4.26,-2.59)--(-3.6463196618288167,-1.2370957722321034), linewidth(0.7) + sexdts); /* dots and labels */ dot((-6.78,4.95),dotstyle); label("$B$", (-6.714545454545459,5.139090909090909), NE * labelscalefactor); dot((-6.78,-2.59),dotstyle); label("$A$", (-7.332727272727277,-3.1518181818181823), NE * labelscalefactor); dot((4.26,-2.59),dotstyle); label("$C$", (4.34,-2.4063636363636367), NE * labelscalefactor); dot((-4.570314034778992,-2.59),dotstyle); label("$D$", (-4.423636363636369,-3.17), NE * labelscalefactor); dot((-2.7101665988177412,-1.3972873014084721),dotstyle); label("$E$", (-2.5509090909090975,-1.97), NE * labelscalefactor); dot((-3.6463196618288167,-1.2370957722321034),linewidth(4.pt) + dotstyle); label("$F$", (-3.86,-0.9518181818181822), NE * labelscalefactor); dot((-0.01585517068551292,0.3302851437471739),linewidth(4.pt) + dotstyle); label("$X$", (0.12181818181817432,0.5209090909090904), NE * labelscalefactor); dot((-1.761560644923938,1.5225513824933414),linewidth(4.pt) + dotstyle); label("$G$", (-1.6963636363636432,1.6663636363636358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Here's a projective solution: Let $G$ be the foot of perpendicular from $D$ to $BC$ and let $X$ be the intersection of $DE$ with $BC$. Let $\omega$ be the circle with diameter $BD$. Then $A \in \omega$, and therefore the reflection of $A$ across $BD$ is also on $\omega$, and $(B,D; A,E) = -1$. Now, $G \in \omega$ therefore $(GB, GD; GA, GE) = -1 = (GC, GD; GA, GE)$. Hence, $A,F,X$ are collinear by Ceva and Menelaus Theorem in $\displaystyle \triangle DGC$.