Triangles $ABC$ and $DEF$ share circumcircle $\Omega$ and incircle $\omega$ so that points $A,F,B,D,C,$ and $E$ occur in this order along $\Omega$. Let $\Delta_A$ be the triangle formed by lines $AB,AC,$ and $EF,$ and define triangles $\Delta_B, \Delta_C, \ldots, \Delta_F$ similarly. Furthermore, let $\Omega_A$ and $\omega_A$ be the circumcircle and incircle of triangle $\Delta_A$, respectively, and define circles $\Omega_B, \omega_B, \ldots, \Omega_F, \omega_F$ similarly. (a) Prove that the two common external tangents to circles $\Omega_A$ and $\Omega_D$ and the two common external tangents to $\omega_A$ and $\omega_D$ are either concurrent or pairwise parallel. (b) Suppose that these four lines meet at point $T_A$, and define points $T_B$ and $T_C$ similarly. Prove that points $T_A,T_B$, and $T_C$ are collinear. Nikolai Beluhov
Problem
Source: TSTST 2021/6
Tags: USA TSTST, geometry
13.12.2021 20:02
Really good geometry problem, appropriate for its placement. Or maybe I just like the techniques used. 6a. Note that the intersection of the two common external tangent to two circles is the center of the unique positive homothety mapping one circle to the other. Let $f_A$ be the positive homothety mapping $\Omega_A$ to $\Omega$, and define $f_D$ similarly. It suffices to show that $f_A$ and $f_D$ both map $\omega_A$ and $\omega_D$, respectively, to the same circle. Let $\triangle{A'B'C'}$ be the image of $\Delta_A$ under $f_A$ and define $\triangle{D'E'F'}$ similarly. Let their incenters be $J_A$ and $J_D$. Let $BC$ intersect $EF$ at $X$. Let $l = XI$ be the interior angle bisector of these two lines. Also let $X'$ be the circumecenter of $\triangle{AID}$. Our first claim is that $X'$ lies on $l$. This follows since if we let $l'$ be the interior angle bisector of $BF$ and $CE$, then angle chasing immediately shows that $l \perp l'$ but also that $l'$ is parallel to $A_1D_1$ where $A_1$ and $D_1$ are the midpoints of "minor" arcs $\overset{\huge\frown}{BC}$ and $\overset{\huge\frown}{EF}$ respectively. This shows $A_1D_1 \perp l$. But we also know $AA_1$ intersects $DD_1$ at $I$ and that from isogonal conjugates (read: angle chasing) in similar $\triangle{A_1ID_1}$ and $\triangle{AID}$ that $A_1D_1 \perp l$ implies that $X'$ lies on $l$ as claimed. Define $B_1$ and its variants in a similar fashion. We have from $CA || C'A'$ that $B_1$ is still the "minor" arc midpoint of $\overset{\huge\frown}{C'A'}$. As a combined corollary, $B_1C_1$ is the perpendicular bisector of $A'J_A$. We also already know $B_1C_1$ is the perpendicular bisector of $AI$ and hence passes through $X'$. From $AB || A'B'$ and $CA || C'A'$ we have that $BC' || B'C$ and moreover that both latter lines are parallel to the "interior" angle bisector of $BC$ and $B'C'$. But $B'C' || EF$, so $B'C || XI$. Now create a dummy point $N$ at the intersection of $AI$ and $BC$. We get: \begin{align*}\angle{IAA'} &= 180^{\circ} - \angle{ACA_1} - \angle{ACA'} = 180^{\circ} - \angle{ACA_1} - \angle{B'CB} \\ &= 180^{\circ} - \angle{ACA_1} - \angle{B'CB} = 180^{\circ} - \angle{INX} - \angle{IXN} \\ &= \angle{AIX'} \\ &= \angle{IAX'}\end{align*}So $A'$ also lies on $AX'$ which also gives that $J_A$ lies on $l$. Similarly $J_D$ lies on $l$. Recall that within our angle chase we showed $\overset{\huge\frown}{AA'} = \angle{BXE}$. Similarly we also have $\overset{\huge\frown}{DD'} = \angle{BXE}$. Thus, $IJ_A = AA' = DD' = IJ_D$, which proves $J_A = J_D$. From Poncelet's Porism, since the incircles of $\triangle{A'B'C'}$ and $\triangle{D'E'F'}$ have the same center, they are necessarily identical, which is exactly what we wanted to show. $\blacksquare$ 6b. Suppose that the common tangents to $\omega_A$ and $\omega_D$ meet at $P$. Let $I_A$ be the incenter of $\Delta_A$ and define $I_D$ similarly. In the spirit of 6a, let $f'_A$ be the positive homothety mapping $\omega_A$ to $\omega$, and define $f'_D$ similarly. They have centers at $A$ and $D$ respectively. Let the image of $\Delta_A$ under $f'_A$ be $\triangle{AB^*C^*}$ and define $\triangle{DE^*F^*}$ similarly. From 6a, we are already given that these share a circumcircle, $\Gamma$. Let $A_1^*$ be the intersection of $AI$ with $\Gamma$ and define $D_1^*$ similarly. Also let $I_A^*$ be the image of the $A$-excenter of $\Delta$ under $f'_A$ and define $I_D^*$ similarly. We know that $A_1^*$ is the midpoint of $II_A^*$. Define $f(X) = \text{Pow}(X, \omega) - \text{Pow}(X, \Omega)$. Linearity of Power of a Point states this function is linear. We have $\frac{I_AP}{I_DP} = \frac{r_A}{r_D} = \frac{r_A}{r} : \frac{r_D}{r} = \frac{AI_A}{AI} : \frac{DI_D}{DI}$ so after applying Menelaus twice we have that $P$ lies on $AD$ and $\frac{AP}{DP} = \frac{AI_A}{I_AI} : \frac{DI_D}{I_DI}$. So now we compute: \begin{align*} F(P) &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot F(A) - \frac{I_DI}{DI_D} \cdot F(D)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot \text{Pow}(A, \omega) - \frac{I_DI}{DI_D} \cdot \text{Pow}(D, \omega)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot (AI^2 - r^2) - \frac{I_DI}{DI_D} \cdot (DI^2 - r^2)\right) \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{I_AI}{AI_A} \cdot AI^2 - \frac{I_DI}{DI_D} \cdot DI^2\right) - r^2 \\ &= \frac{1}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\frac{II_A^*}{AI} \cdot AI^2 - \frac{II_D^*}{DI} \cdot DI^2\right) - r^2 \\ &= \frac{2}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(IA_1^* \cdot AI - ID_1^* \cdot DI\right) - r^2 \\ &= \frac{2}{\frac{I_AI}{AI_A} - \frac{I_DI}{DI_D}} \left(\text{Pow}(I, \Gamma) - \text{Pow}(I, \Gamma)\right) - r^2 \\ &= -r^2 \end{align*}The locus of points where $F$ is constant is necessarily a line, so we are done. $\blacksquare$
13.12.2021 20:25
Part (a) is implied by the following, which I haven't been able to prove: Quote: Let $B', C'$ be the points on $AB, AC$ such that $\overline{B'C'}$ is parallel to $\overline{EF}$ and tangent to $\omega$ (i.e. $\triangle AB'C'$ is the image of $\Delta_A$ under the homothety at $A$ sending $\omega_A$ to $\omega$). Similarly define $E',F'$. Then $AB'C'DE'F'$ are concyclic. can someone help? (amusingly, this statement is its own converse, if we consider $AB'C'DE'F'$ as the original hexagon)
13.12.2021 20:46
@above maybe something like the following: Construct B’, C’ instead such that B’ lies on AB, C’ lies on AC, $\angle EDC = \angle FDB’$, and $\angle FDB = \angle EDC’$. By Pascal B’C’ is parallel to EF. The goal is to show this is the same definition as yours, i.e. B’C’ is tangent to $\omega$. Let X = B’C’ cap BC. We use DDIT on the point D and the quadrilateral ABXC’ and an inconic tangent to its four sides plus the side DE. (This inconic exists except when three of the five sides concur; if that happens, replace DE with DF and DF with DE in the argument below.) Then the involution is just “reflection across internal angle bisector of EDF”, so DF is tangent to the said inconic as well, so this inconic is just the incircle $\omega$, which shows that B’C’ is tangent to it. The rest is just angle chasing to prove AB’C’D concyclic (D is the Miquel point of ABXC'). A similar argument shows AE’F’D concyclic; finally, because $\angle F'AB' = \angle EAC = \angle EDC = \angle F'DB'$, the six points lie on the same circle.
14.12.2021 04:52
There was a lot of work put into the official solutions for this problem. I hope you enjoy some of them. Let $I$ and $r$ be the center and radius of $\omega$, and let $O$ and $R$ be the center and radius of $\Omega$. Let $O_A$ and $I_A$ be the circumcenter and incenter of triangle $\Delta_A$, and define $O_B$, $I_B, \dotsc, I_F$ similarly. Let $\omega$ touch $EF$ at $A_1$, and define $B_1$, $C_1, \dotsc, F_1$ similarly.
14.12.2021 09:27
the vertices of $\Delta_A$ are $A,X_A,Y_A$ in clockwise order, $O_A$ and $I_A$ are the circumcenter and incenter of $\Delta_A$ resp. denote the center of mixtilinear incircle associated with $A$ and $\triangle ABC$ with $M_A$. Part a:
Let $P=AD' \cap A'D$. It is obvious that $P$ is fixed (since in (1) we already proved that $(A',D')$ and $(A,D)$ are anti-homologous pairs) so $T_A$ is on the polar of $P$ wrt $\Omega$. $\blacksquare$
[1] https://artofproblemsolving.com/community/q3h186117p1022749
24.04.2022 09:26
27.08.2022 21:21
Great problem! It took me a lot of time but it worthwhile! First notice that by Poncelet's Porism, we can take $D$ as any point. Now make some construction to caracterize the extimilicenter of $(\Omega_A,\Omega_D).$ Let $E'\in DE, F'\in DF$ such that $E'F'\parallel BC$ and $E'F'$ is tangent to $\omega$ (i.e. the intersections of the reflection of $BC$ over $I.$ Similarly, define points $B'$ and $C'.$ Let $I$ be the center of $\omega.$ Claim 1: Points $A,E',F'$ and $D'$ are concyclic. Proof: Notice that since $EE'$ and $F'F$ concurr at $D$ and $A \in (DEF),$ our claim is equivalent to show that $A$ is the Miquel Point of lines $\overline{DE'E}, \overline{DFF'},\overline{EF}$ and $\overline{E'F'}.$ Therefore, if we let $H:=EF \cap E'F',$ it suffies to show that points $AHEF'$ are concyclic. In other words, we have to show that $$\measuredangle (AH,AE) =\measuredangle (F'H,F'F) = \measuredangle (E'F', FD) = \measuredangle (BC,FD) = $$$$=\measuredangle CAF + \measuredangle BAD $$On the other hand $$ \measuredangle (AH,AF) = \measuredangle HAC + \measuredangle CAF $$So we need to show that $\measuredangle BAD = \measuredangle HAC,$ i.e., that $AH$ and $AD$ are isogonal conjugate with respect to $\angle BAC.$ Now we use unheterred moving points. Fix$\triangle ABC$ and let $S_A$ be the tangency point of the $A$-mixtilinear incircle with $\Omega$ and let $X_A$ be the tangency point of $\omega$ with $\overline{BC}.$ Animate point $D$ over $\Omega,$ and let $Q$ be the second intersection of $DX_A$ with $\Omega$ and let $P:=\overline{EF} \cap \overline{BC}.$ First notice that by DDIT $P,Q$ and $A$ are colinear. Then $D$ has degree 2, then $P$ also has degree $2$ (since the map $D\rightarrow P$ is projective); then $Q$ has degree $2$ and therefore line $EF$ (the reflection of $\overline{BC}$ over $\overline{DI}$ has degree $2.$ Therefore $H$ has degree at most $2.$ On the other hand the isogonal of $AD$ has degree $1.$ Then we only have to check that $H$ lies on the isogonal conjugate of $AD$ for $4$ choices of $D.$ The degenerate cases $D=A,B,C$ are trivial; to finish, consider the case $D=S_A,$ then $\overline{EF}\parallel \overline{BC}$ (since $\overline{S_A}$ passes through the arc midpoint of $BAC$) and $H$ is the reflection of $X_A$ over $I;$ thus indeed $AS_A$ and $AH$ are isogonal due to the well known facts that $AH$ passes through $Y_A,$ the tangency point of the $A$-excircle with $\overline{BC},$ and the fact that $AT_A$ and $AY_A$ are isogonal. $\square.$ Similarly we show that $A,B',C',D$ are concyclic. Further, let $K:=BC\cap B'C'$ and let $D'$ and $A'$ be the second intersections of lines $AH$ and $DK,$ respectively. Then, we have that $D'$ is the reflection of $D$ over the perpendicualr bisector of $BC$ and similarly $A'$ is the reflection of $A$ over the perpendicular bisector of $EF.$ Now let $O_A$ and $O_D$ be the centers of $\Omega_A$ and $\Omega_D,$ respectively. We then claim the following Claim 2: Lines $AO_A,DO_D$ and the perpendicuar bisector of $AD$ concurr. Proof: Our claim is equivalent to show that $$\measuredangle (AD,AO_A) = \measuredangle (DO_D,AD)$$Fist note that since $AD$ and $AH$ are isogonal conjugate with respect to $\angle BAC$ and $AO_A$ is isogonal to the perpendicular from $A$ to $EF$ with respect to $\angle BAC,$ we have that $$\measuredangle (AD,AO_A) = 90 + \measuredangle (EF, AH) = 90 + \measuredangle = 90 + \measuredangle BAF + \measuredangle D'AB + \measuredangle ABE $$Similarly, $$\measuredangle (DO_D, AD) = 90 + \measuredangle CBD + \measuredangle CDA' + \measuredangle A'DA$$Therefore it suffies to show $$\measuredangle BAF + \measuredangle D'AB + \measuredangle ABE = \measuredangle CBD + \measuredangle BDF + \measuredangle FDA'$$which is obvious by the isogonal definition of $A'$ and $D'.~ \square$ Now let $R$ be the intersection of $AO_A, DO_D$ and the perpendicular bisector of $AD.$ Considering the homothety with center $A$ sending ${\triangle}_A$ to $\triangle AB'C',$ we must have that the center of $(AB'C')$ must lie on $AO_A$ and at the perpendicular bisector of $AD,$ (by the homothety and by Claim 1) therefore, $R$ is the center of $(AB'C'D).$ Analogously, we have that $R$ is the center of $(AE'F'D)$ and then the six points $A,D,B',C',E',F'$ must lie on a circle centered at $R.$ Considering the homothetys again we can state the following: Lemma wrote: Let $I_A$ and $I_D$ are the centers of ${\omega}_A$ and ${\omega}_D.$ We have that $(i)$ The points $A,D,B',C',E',F'$ lie on a circle ${\Gamma}_A$ which is tangent to ${\omega}_A$ at $A$ and to ${\omega}_D$ at $D.$ $(ii)$ The lines $O_AI_A, RI$ and $O_DI_D$ are pairwise parallel (equivalently, they concurr at the line at infinity). Now we are able to establish part (a); apply Monge's Theorem to $({\Omega}_A, {\Omega}_D, {\Gamma}_A)$ and $({\omega}_A, {\omega}_D, \omega)$ to conclude that the extimilicenters of $({\Omega}_A,{\Omega}_D)$ and $({\omega}_A,{\omega}_D)$ both lies on $AD.$ Then, it suffies to show that lines $O_AO_D,I_AI_D$ and $AD$ concurr; indeed, this follows from part $(ii)$ of the Lemma applying Desargue's Theorem to $\triangle AO_AI_A$ and $\triangle DO_DI_D$. Now we have that $T_A$ lies on $AD;$ we need to identify a second line. Let $S_D$ be the tangency point of the $D$-mixtilinear incircle of $\triangle DEF.$ We then have the following: Claim 3: $T_A$ lies on $S_AS_D.$ Proof: The key idea is to notice that the line $E'F'$ is the line passing through the intersections of $\Omega$ with the tangents from $S_A$ to $\omega;$ then we can find nice relations by swaping $A\leftrightarrow S_A$ and $D\leftrightarrow S_D$ and apply the Lemma. In which follows, dentoe by $\triangle (P,Q)$ the corresponding of $\Omega_A$ by swapping $A\leftrightarrow P$ and $D \leftrightarrow Q$ for any points $P,Q \in \Omega$ and further let $\Omega (P,Q)$ be the circumcircle of $\triangle P,Q$ (note that this is posible by Poncelet's Porism; for intance we have $\triangle (D,A)={\triangle}_A$.) Now consider the configuration generated by $D \leftrightarrow S_D; $ then it follows by applying the part (i) of the Lemma that $\Omega (S_D,S_A) = \Omega (A,D);$ similarly, we have that $\Omega (S_A,S_D) = \Omega (D,A).$ Then from applying the part (a) of the problem to the configuration $(A,D) \leftrightarrow (S_D,S_A)$ we have that the extimilincenter of $(\Omega (S_A,S_D), \Omega (S_D,S_A))$ lies on $S_AS_D;$ but by we have just shown that $(\Omega (S_A,S_D), \Omega (S_D,S_A))= ({\Omega}_D, {\Omega}_A). \square.$ Then we finally have that $T_A = AD \cap S_AS_D$ (the case $D=S_A$ is done by continuity). To finish, let $P:=AT_A \cap DT_D;$ therefore $T_A$ lies on the polar of $P$ with respect to $\Omega.$ Now notice that $P$ is a point not depending at $A$ or $D$ (i.e. it can be determined uniquely by $\omega$ and $\Omega$). Indeed, by applying Monge's Theorem to the incircle, circumcircle and mixtilinear incircle, $P$ is the extimilincenter of $\omega$ and $\Omega.$ Another way to see that $P$ is fixed, is to notice that the map $A\Rightarrow S_A$ is projective and furthermore an involution (since $S_A$ can be defined as the point such that the second intersection of $AI$ and $S_AI$ cut $\Omega$ in diametrically oppositive points), and therefore $AS_A$ must pass through a fixed point.
17.05.2023 23:04
Cant believe this geo was once something i called impossible and now here i'm :O, i'm pretty sure this is one of the best geos i've ever seen in my whole life so far (May 17th of 2023), congrats to Nikolai Beluhov for making such a good problem . Let $B',C'$ points in $AB,AC$ respectivily such that $B'C' \parallel EF$ and $B'C'$ tangent to $\omega$, let $E'F'$ points in $DE,DF$ respectivily such that $E'F' \parallel BC$ and $E'F'$ tangent to $\omega$, now let $DE$ hit $BC,AC$ at $D_2,E_2$ respectivily and let $DF$ hit $AB,BC$ at $F_2,D_1$ respectivily and finally let $EF$ hit $AB,AC$ at $F_1,E_1$ respectivily. Now let $\Omega_A \cap \Omega=A'$ and $\Omega_D \cap \Omega=D'$. Claim 1: $(\Omega_A \cap \omega_A) \sim (\Omega_D \cap \omega_D)$ (this implies part a) Proof: Let $D''$ the miquel point of complete quad $\{B'C',BC, AB, AC \}$, now clearly $\angle BD''B'=\angle (BC, B'C')=\angle CD''C'$ so $D''B, D''C'$ is isogonal w.r.t. $\angle B'D''C$, now by DDIT over that complete quad and inconic $\omega$ we get that $D''I$ bisects $\angle B'D''C, \angle BD''C'$, let $\ell_{D''}$ the tangent from $D''$ to $AB'D''C'$, we have $\angle (\ell_{D''}, D''B')=\angle B'C'D''=\angle BCD''= \angle CD''\infty_{E'F'}$ so $D''I$ bisects $\angle E'DF'$ as well this means that $D''=D$ so $AB'DC'$ is cyclic, now clearly $(AB'DC')$ is tangent to $\Omega_A$ but we can also show it is tangent to $\Omega_D$, notice that this is equivalent to $\angle D_1D_2D-\angle B'DC'=\angle D_1DB'$ but this holds as $\angle D_1D_2D-\angle B'DC'=\angle BD_2D-\angle BDC=\angle EDC=\angle D_1DB'$ thus $(AB'DC')$ is tangent to $\Omega_D$ as well, now notice that by the same process u get that $AE'DF'$ is cyclic and $(AE'DF')$ tangent to both $\Omega_A$ and $\Omega_D$ so they are basically the same circle, hence $AF'B'DC'E'$ is cyclic and $(AF'B'DC'E')$ tangent to $\Omega_A$ and $\Omega_D$ so now to finish we have $(\Omega_A \cap \omega_A) \sim ((AF'B'DC'E') \cap \omega) \sim (\Omega_D \cap \omega_D)$ thus we are done. Claim 2: $T_A, T_B, T_C$ all lie in the polar of $X_{57}$ w.r.t. $\Omega$ (implies part b) Proof: Notice that here $\triangle ABC$ and $\triangle DEF$ share same $X_{57}$, now by Taiwan TST 2014 R3 P3 we have that $D'$ is the A-mixtilinear intouch point of $\triangle ABC$ and $A'$ is the D-mixtilinear intouch point of $\triangle DEF$, by monge over $\Omega_A, \Omega_D, (AF'B'DC'E')$ we have that $AD$ passes through $T_A$ so make a $\sqrt{T_AA \cdot T_AD}$ inversion with center $T_A$, this sends $\Omega_A \to \Omega_D$ but since $ADD'A'$ is cyclic we have that $A' \to D'$ so $A'D' \cap AD=T_A$, now by monge its known that $AD' \cap DA'=X_{57}$ so by Brokard over $\Omega$ we have that $T_A$ lies in the polar of $X_{57}$, but now u can do the same for all $T_B, T_C$ so $T_A,T_B,T_C$ make a line which is the polar of $X_{57}$ w.r.t. $\Omega$ thus we are done . Wow, what a journey, waiting for more problems like this coming soon .
18.05.2023 15:11
Denote with $O_\bullet$ and $I_\bullet$ the centers of $\Omega_\bullet$ and $\omega_\bullet$, respectively. We first prove the following problem: Lemma wrote: Let $\triangle ABC$ have incircle $\omega$ and let $P$ be some point on the circumcircle $\Omega$. If $\triangle PQR$ is the triangle which has incircle $\omega$ and $QR\parallel BC$ prove that $APQR$ is cyclic.
Now we begin with some Claims. Claim 1: $A$ and $D$ are antihomologous points w.r.t. $\Omega_A$ and $\Omega_D$. Proof. Let $AD\cap\Omega_A=\left\{A,D'\right\}$, it then suffices $\angle AD'O_A=\angle ADO_D$ \[\Longleftrightarrow \angle EDA-\angle EDO_D=\angle CAD-\angle CAO_A\Longleftrightarrow \angle EDA-\angle CAD=\angle ABC-\angle DFE\]which is seen to be correct. Claim 2: $\left\{\Omega_A\cup\omega_A\right\}\sim\left\{\Omega_D\cup\omega_D\right\}$ which implies a). Proof. Taking the homothety centered at $A$, $\chi:\omega_A\rightarrow\omega$ we get from the lemma that $D\in\chi(\Omega_A)$ and hence $\chi(D')=D\Longrightarrow D'I_A\parallel DI_D$ and similarly $A'I_D\parallel AI_A$ where $A'$ is analogously defined. Let $T_A$ be the homothetic center of $\triangle AD'I_A\sim\triangle A'DI_D$ and we see that $T_A$ is the exsimilicenter of $\Omega_A$ and $\Omega_D$ and by Monge, the exsimilicenter of $\omega_A$ and $\omega_D$ is on $AD$ hence is $T_A$. From the Taiwan TST 2014 problem we know that $S_A$, the $A$-mixtilinear intouch point of $\triangle ABC$, is on $\Omega_D$ and similarly $S_D\in\Omega_A$. Claim 3: $S_A$ and $S_D$ are antihomologous w.r.t. $T_A$. Proof. This is true because of Claim 1 and the fact that pairs of antihomologous points form cyclic quadrilaterals (here $AS_DS_AD$ is cyclic). And for the final blow: $T_A$ is on the polar of the exsimilicenter $X$ of $\Omega$ and $\omega$ w.r.t. $\Omega$. This is due Monge $X=AS_A\cap DS_D$ and by Claims 1 and 3, $T_A=S_AS_D\cap AD$ hence Brokard's theorem on $AS_DS_AD$ finishes.