My #200 post .... Let $\omega$ the incircle with center $I$ and $P$, $Q$, $R$ the points of tangency of $\omega$ with $BC$, $CA$, $AB$. Let $P'=PI\cap \omega$ and $M=AP'\cap BC$ and $N=AP'\cap \omega$, since $MC=BP$(well-know) $\Longrightarrow$ $PD=DM$ and $\measuredangle PNM=90^{\circ}$ $\Longrightarrow$ $DP=DM=DN$ hence $DN$ is tangent to $\omega$. Let $S=DN\cap AC$ and define $BP=BR=2a$, $CP=CQ=2b$, $AR=AQ=2c$, $QS=x$ $\Longrightarrow$ $DP$ $=$ $DM$ $=$ $DN$ $=$ $b-a$. $\Longrightarrow$ by Menelao's theorem in $\triangle DSC$ with line $\overline{ANM}$ we get: $$CM.DN.SA=DM.NS.AC \Longrightarrow 2a.(b-a).(2c+x)=(b-a).x.(2b+2c)$$$$\Longrightarrow 2a.(2c+x)=x.(2b+2c) \Longrightarrow x=\tfrac{2ac}{b+c-a}$$On the other hand: $\tfrac{XE}{XF}=\tfrac{XE}{XE+MC}=\tfrac{ES}{EC}$, since $SC=2b-x=b+c-ES$ we get $ES=\tfrac{(b+c)(a+c-b)}{b+c-a}$ Hence $\tfrac{XE}{XF}=\tfrac{ES}{EC}=\tfrac{c+a-b}{c+b-a}$ similarly $\tfrac{YF}{YD}=\tfrac{a+b-c}{a+c-b}$ and $\tfrac{ZD}{ZE}=\tfrac{b+c-a}{b+a-c}$ $\Longrightarrow$ $\tfrac{XE}{XF}.\tfrac{YF}{YD}.\tfrac{ZD}{ZE}=1$ which is the converse of Menelao's theorem in $\triangle DEF$ with line $\overline{XYZ}$ $\Longrightarrow$ $X$, $Y$, $Z$ are collinear.

Can someone post a complex solution. I think we can easily get $X$.

Lemma:In triangle $ABC$ let the excircle touch $AB,AC$ in $P,Q$.Let $R,S$ be the midpoints of $AP,AQ$ and let $X=\{RS\}\cap BC$.Similarly define $Y,Z$ than X,Y,Z are collinear. Proof: It's immediate by Menelaus on RS as a transversal. Let $P'$ be the touch point of $A$-excircle $P$ of the incircle and let $(PP')\cap \odot I=\{R\}$.Now as $DP$ is tangent to $\odot I$ so is $DR$ ,$P'R\perp PR$ as well as $ID\perp PR$ and so $ID||P'R$ and hence as $ID$ is the A-Nagell cevian in the medial triangle so is $AR$ in $\triangle ABC$.Let $DI\cap EF=\{V\}$, now $\angle XVD=\frac{\angle PDR}{2}=\angle PP'R=\angle VDX$ $\implies$ $\triangle VDX$ is $X$-isosceles.By previous,$X$ is on the perpendicular bisector of $DV$ and so on the radical axis of point $D$ and $D$-exicrcle.Now applying the lemma on $\triangle DEF$ we're done.$\blacksquare$

Easy with Barycentre.

Should be easy to write the solution than.

Murad.Aghazade wrote: Easy with Barycentre. How tangents condition makes it quite ugly

My solution. Let N be the intersection of the reflection of $BC$ in $I$, the incenter and the line $DX$. By the dual of desargues involution theorem, the involution mapping DN to DB and the line joining D to the incircle touch point with AB to the line joining D to the antipode of the incircle touchpt with BC in the incircle, fixes the intersection of the reflection of BC in I with AB. Now just use preservation of cross ratios and done.

Let $ \triangle I_aI_bI_c $ be the intouch triangle of $ \triangle ABC $ and let $ U $ be the intersection of $ CA, $ $ DX. $ By Newton's theorem we get $ AD, $ $ BU, $ $ I_aI_b $ are concurrent, so $ (A,C;I_b,U) $ $ = $ $ (D,C;I_a,B) (\bigstar). $ Let $ T $ be the Nagel point of $ \triangle ABC $ and let $ V $ be the intersection of $ CA, $ $ DT. $ It's well-known that the reflection $ \widetilde{I_a}, $ $ \widetilde{I_b} $ of $ I_a, $ $ I_b $ in $ D, $ $ E $ lies on $ AT, $ $ BT, $ respectively, so $$ (A,C;I_b,U) \stackrel{(\bigstar)}{=} \underbrace{(D,C;I_a,B) = (D,B;\widetilde{I_a},C)}_{\text{isotomic conjugate WRT B and C}} \stackrel{T}{=} (C,A;\widetilde{I_b},V) \Longrightarrow EU = EV \ , $$hence we conclude that $ D(E,F;X,T) $ $ = $ $ -1 $ $ \Longrightarrow $ $ X $ lies on the trilinear polar of $ T $ WRT $ \triangle DEF. $

Describe the position of the tangent point in another way and use Menelaus，we can solve this problem easily.

Let $XYZ$ be the triangle formed by the three tangents. By Desargues' Theorem, it suffices to prove that $DX$, $EY$, and $FZ$ concur. We claim that they all pass through the Nagel point $Na$ of $ABC$. Let $ABC$ have intouch triangle $PQR$ and incenter $I$, and let $P'Q'R'$ be the reflection of $PQR$ over $I$. Let $AP'$, $BQ'$, and $CR'$ intersect the incircle again at $U$, $V$, and $W$. It is well known that $P'U$, $Q'V$, and $R'W$ concur at $Na$ an that $DU$, $EV$, and $FW$ are tangents to the incircle. Note that the tangents to the incircle at $V$ and $W$ intersect at $X$. Let the tangents to the incircle at $Q'$ and $R'$ intersect at $X'$. Note that as $Q'V$ and $R'W$ intersect at $Na$, $Na$ lies on $XX'$. We will show that $D$ lies on $NaX'$, which will imply the desired. Note that $X'$ is the reflection of $A$ over $I$. Hence, if $G$ is the centroid of $ABC$, it is also the centroid of $NaAX'$ as $G$ divides median $NaI$ in the ratio $2:1$. Thus, the point $D$ on $AG$ with $AG:GD=2:1$ must be the midpoint of $X'Na$ and we are done.

Let $H_A$ be the orthocenter of $\triangle BIC$, and define $H_B$, $H_C$ similarly. Suppose $I$ is the incenter, $R$ is the intouch point on $\overline{BC}$, and let $S\equiv DX\cap \odot(I)$; I claim that $H_AS$ is concurrent with its cyclic variations. To prove this, since $H_AR$ and cyclic variations are obviously concurrent at $I$, we have $$1=\prod_{\text{cyc}}\frac{H_BR\cdot H_BS}{H_CR\cdot H_CS}=\prod_{\text{cyc}}\frac{H_BS}{H_CS}$$where the first equality follows from Power of a Point. Call this concurrency point $M$; notice that $X$ lies on the polars of $H_A$ and $S$ with respect to $\odot(I)$, so $H_AS$ is the polar of $X$ with respect to $\odot(I)$. Therefore, $\overline{XYZ}$ is a line segment on the polar of $M$ with respect to $\odot(I)$, and the conclusion follows.

Thanks for pointing out my mistake. Ignore this post.

Complex2Liu wrote: Let the incircle of $ABC$ touch $BC$ at $P,$ and $P_1,P_2$ be the reflection of $P$ with respect to $I,D$ respectively. Denote by $U$ the second intersection of $AP_2$ and $\odot(I).$ It's well-known that $A,P_1,P_2$ are collinear and $ID \parallel AP_2,$ from which we conclude that $DU$ is tangent to $\odot(I).$ Now let $V$ be the intersection of $EF$ and $AU,$ and $X^*$ be the reflection of $V$ with respect to $X.$ Notice that \[\measuredangle XVU=\measuredangle DP_2U=\measuredangle P_2UD=\measuredangle VUX \implies XV=XU. \]Hence $XV\cdot XX^*=XU^2=XF\cdot XE,$ which amounts to $(X^*,V;F,E)\stackrel{U}{=}-1\implies UX$ is the angle bisector of $\angle FUE.$ Therefore $\tfrac{XF}{XE}=\tfrac{VF}{VE}=\tfrac{BP_2}{CP_2}.$ The statement then follows by Menelaus' theorem in $\triangle DEF$(It's well-known that three lines are concurrent at Nagel point). [asy][asy] size(8cm); defaultpen(fontsize(9pt)); pathpen=black; pointpen=black; pair A=dir(55); pair B=dir(-140); pair C=dir(-40); pair D=midpoint(B--C); pair E=midpoint(A--C); pair F=midpoint(A--B); pair I=incenter(A,B,C); pair P=foot(I,B,C); pair P2=2*D-P; pair P1=2*I-P; pair U=OP(incircle(A,B,C),P1--P2); pair V=IP(E--F,P1--U); pair X=extension(E,F,D,U); pair X1=2*X-V; D(A--B--C--cycle,purple+linewidth(1.1)); D(incircle(A,B,C),dotted+red); D(A--P2,cyan); D(D--X,dashed); D(X1--E,red+linewidth(1.2)); D(X1--P--P1); D(E--U--F); D(rightanglemark(P,U,P1,2),deepgreen); 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("$I$",I,dir(I)); dot("$P$",P,dir(P)); dot("$P_1$",P1,dir(120)); dot("$P_2$",P2,dir(P2)); dot("$U$",U,dir(-160)); dot("$V$",V,dir(135)); dot("$X$",X,dir(90)); dot("$X^*$",X1,dir(90)); [/asy][/asy] Warning: $E,F$ are not on the incircle, so be aware of the fact that the property $XU^2=XE \cdot XF $ does not hold.

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But how can we prove that ? It is difficult to think of that, as well.

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I still do not know how to prove it. How can we relate that to X point ?

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Equivalent form : Posted here

MNZ2000 wrote: There is very shorter solution: Consider polar with center at $I$ and ratio $r^2$. since $X$ is on the $EF$ so its on the polar of $A$. If $P$ is point such $DP$ is tangent to incircle, $X$ is also on the polar of $P$ so pole of $X$ is $AX$ lemma: $AX$ passes through tangent point of A-exircle with $BC$ (call this point $D'$ define $E',F'$ similary. Proof is easy We shold prove polar of $X,Y,Z$ are concurent. Means prove $AD',BF',CE'$ are concurrent And its obvious You think $EF$ is the polar of $A$ w.r.t. the incircle?? No it's not.

ABCDE wrote: Let $XYZ$ be the triangle formed by the three tangents. By Desargues' Theorem, it suffices to prove that $DX$, $EY$, and $FZ$ concur. Sorry but it is not always that $DX,EY,FZ$ concur.

Let $(I)$ be the incircle of $\triangle ABC$ and let $L, M, N$ be the touchpoints of $(I)$ with $BC, CA, AB$, respectively. Now let $L'$ be the antipode of $L$ in $(I)$ and $L_1$ be the reflection of $L$ WRT $D$. Suppose, $AL \cap EF = L_2, AL_1 \cap L_3, AD \cap EF = D'$. $\triangle AEF$ and $\triangle ACB$ are homothetic with homothety center $A$. So, the homothety with center $A$ and ratio $\dfrac {1} {2}$ sends $L$ to $L_2$, $L_1$ to $L_3$ and $D$ to $D'$. So, $L_2$ is the touchpoint of the incircle of $\triangle AEF$ with $EF$. Now, $\triangle AEF$ and $\triangle DFE$ are also homothetic with homothety center $D'$ and ratio $-1$. Since , $D'L_2 = D'L_3$, so, $L_3$ is the touchpoint of the incircle (call it $(I')$) of $\triangle DEF$ with $EF$. Now, it is well known that $A, L', L_1$ are collinear. Now let $AL_1 \cap (I) = L_4$. Then $\angle LL_4L_1 = 90^{\circ}$. So, $DL = DL_4$ and so $L_4$ is the touchpoint of the tangent $DX$ with $(I)$. Now $\angle XL_3L_4 = \angle L_4L_1D = \angle DL_4L_1 = \angle XL_4L_3$ $\Rightarrow$ $XL_3 = XL_4$. So, $X$ lies on the radical axis of $(I)$ and $(I')$. Similarly, $Y$ and $Z$ also lie on the radical axis. So $X, Y, Z$ are collinear.

My solution: Let $M,N$ be the $A$-intouch and $A$-extouch points in $\triangle ABC$. Also let $DX$ meet the incircle at $T$, and $M'$ be the antipode of $M$ in the incircle. From here, we get that $M'$ and $T$ lie on $AN$, and that $D$ is the center of $\odot (MTN)$. Let $AN \cap EF = S$. Then by the homothety centered at $A$ that sends $\triangle AEF$ to $\triangle ABC$, we get that $S$ is the $A$-extouch point in $\triangle AEF$. As $AEDF$ is a parallelogram, we have that $S$ is the $D$-intouch point in $\triangle DEF$. Now, $\angle XST=\angle DNT=\angle DTN=\angle XTS \Rightarrow XS=XT \Rightarrow X$ lies on the radical axis of the incircles of $\triangle DEF$ and $\triangle ABC$. Similarly, $Y$ and $Z$ lie on the radical axis of the incircles of $\triangle DEF$ and $\triangle ABC$. Thus, $X,Y,Z$ are collinear. $\blacksquare$

here's a different approach using poles and polars: let $R,S,T$ the tangency points and $R',S',T'$ the second tangency points from $D,E,F$ and let $X'=SS' \cap TT'$ difine $Y',Z'$ similary $polar(E)=TT'$ $polar(F)=RR' \implies polar (X)=X'R'$ so $X,Y,Z $ are collinear is equivalent to $X'R',Y'S',Z'T'$ are concurrent which is equivalent to $X'R,Y'S,Z'T$ are concurrent $polar(EF \cap BC ) =X'R $ but $EF \cap BC =P_{\infty BC}$ thus $I \in X'R$ similary $I \in Y'S$ $I \in Z'T$ and we win

Quite a simple problem; yet it took me quite a while to realize the solution.

tenplusten wrote: Easy with Barycentre. Indeed, easy with barycentric coordinates. [asy][asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair A1 = foot(I,B,C); pair E = (A+C)/2; pair F = (A+B)/2; pair D = (B+C)/2; path inc = incircle(A,B,C); pair A4 = B+C-A1; pair A3 = foot(A1,A,A4); pair A2 = 2*I-A1; pair X = extension(D,A3,F,E); draw(A--B--C--A--cycle); draw(inc); draw(E--F); draw(F--X); draw(D--X); draw(A--A4); 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("$I$",I,dir(I)); dot("$X$",X,dir(X)); dot("$A_1$",A1,dir(A1)); dot("$A_2$",A2,dir(A2)); dot("$A_3$",A3,dir(A3)); dot("$A_4$",A4,dir(A4)); [/asy][/asy] Let $A_1$ be the point of tangency of the incircle with side $BC$, and let $A_1 A_2$ be a diameter of the incircle. Let the other tangent from $D$ to the incircle meet the incircle at $A_3$. Then $\angle A_1 A_3 A_2=90^{\circ}$. The extension of $A_2A_3$ meets the side $BC$ at a point $A_4$. It is well-known that $A_4$ is the point of tangency of the $A$-excircle with the side $BC$. Thus, $$A_4=(0:s-b:s-c)$$Note that the displacement vector $\overrightarrow{AA_4}$ is given by $$\left(0,\frac{s-b}{a},\frac{s-c}{a}\right)-(1,0,0)=(-a:s-b:s-c)$$ Next, we use the following fact: Let $(l:m:n)$ be a displacement vector. Then a displacement vector perpendicular to $(l:m:n)$ is given by $$(a^2(n-m)+(c^2-b^2)l:b^2(l-n)+(a^2-c^2)m:c^2(m-l)+(b^2-a^2)n)$$In this case, a displacement vector perpendicular to $AA_4$ is given by $$(2a(b-c)s:-ab^2-b^2(s-c)+(a^2-c^2)(s-b):c^2a+c^2(s-b)+(b^2-a^2)(s-c))$$Simplifying, this is equal to $$(2a(b-c):a^2-2ab-(b-c)^2:-a^2+2ac+(b-c)^2)$$Also, $A_1=(0:s-c:s-b)$. Thus, we can parametrise the line $A_1 A_3$ is $$(2a(b-c)t:s-c+t(a^2-2ab-(b-c)^2):s-b+(-a^2+2ac+(b-c)^2))$$Substituting this into line $AA_4$ given by $-(s-c)y+(s-b)z=0$, we get that $$t=\frac{-b+c}{a^2-ab-ac-2b^2+4bc-2c^2}=\frac{2(b-c)}{a(s-a)+4(b-c)^2}$$From this, $$A_3=(-4(b-c)^2:(a-b)^2-c^2:(c-a)^2-b^2)=(-4(b-c)^2:-4(s-b)(s-a):-4(s-c)(s-a))=\left(\frac{(b-c)^2}{s-a}:s-b:s-c\right)$$Similarly, $$B_3=\left(s-a:\frac{(c-a)^2}{s-b}:s-c\right)$$$$C_3=\left(s-a:s-b:\frac{(a-b)^2}{s-c}\right)$$From this, we get that $DA_3$ is $$(s-a)x+(b-c)y+(c-b)z=0$$Since $EF$ is given by $-x+y+z=0$, we obtain $$X=(b-c:s-c_b-s)$$Similarly, $$Y=(c-s:c-a:s-a)$$$$Z=(s-b:a-s:a-b)$$It remains to verify that $$\begin{vmatrix} b-c & s-c & b-s \\ c-s & c-a & s-a \\ s-b & a-s & a-b \end{vmatrix}=0$$But this is true as the sum of the three rows is the zero row. Thus, $X$, $Y$ and $Z$ are collinear.

Let $I$ be the incenter, the incircle touch $BC, CA, AB$ at $A_1, B_1, C_1$ respectively, the aforementioned tangents from $D, E, F$ touch the incircle at $P_1, P_2, P_3$ respectively, $H_1$ be the orthocenter of $BIC$, $Q = CI \cap BH_1$, and $R = CI \cap EF$. Claim: $BH_1$ is the polar of $R$ (wrt the incircle). Proof. Since $BH_1 \perp CI \equiv IR$, it suffices to show $B$ lies on the polar of $R$. By the Iran Lemma, $A_1C_1$ passes through $R$. But $A_1C_1$ is the polar of $B$, so the desired result follows by La Hire's. $\square$ Claim: $EF$ is the polar of $H_1$. Proof. La Hire's implies $R$ lies on the polar of $H_1$. Because $H_1I \perp BC$ and $EF \parallel BC$, we know $IH_1 \perp EF$. But $R \in EF$, so $EF$ is the polar of $H_1$, as required. $\square$ If we let $H_2, H_3$ be the orthocenters of $CIA, AIB$ respectively, then analogous reasoning implies $FD$ is the polar of $H_2$ and $DE$ is the polar of $H_3$. Claim: $H_1P_1$ is the polar of $X$. Proof. Trivially, $P_1$ lies on the polar of $X$. Since $X \in EF$, i.e. the polar of $H_1$, we know $H_1$ lies on the polar of $X$ by La Hire's. $\square$ Similarly, we conclude $H_2P_2, H_3P_3$ are the polars of $Y, Z$. By the Concurrent Polars Induces Collinear Poles Lemma, it suffices to show $H_1P_1, H_2P_2, H_3P_3$ concur. Claim: $H_1, H_2, C_1, P_3$ are collinear. Proof. By tangency, it's easy to see $C_1P_3$ is the polar of $F$. But we've previously shown that $F$ lies on the polars of $H_1$ and $H_2$, so both orthocenters lie on the polar of $F$ by La Hire's, which suffices. $\square$ Analogously, we conclude $H_2, H_3, A_1, P_1$ and $H_3, H_1, B_1, P_2$ are also sets of collinear points. Now, observe $H_1A_1, H_2B_1, H_3C_1$ concur at $I$. Since $P_1, P_2, P_3$ all lie on $(A_1B_1C_1)$, the desired concurrency follows from the existence of the Cyclocevian Conjugate. $\blacksquare$

Trigbash for the win! Let the incircle $\omega$ touch $BC$ at $P$, $CA$ at $Q$ and $AB$ at $R$. Let $D_2$ be the antipode of $P$ and $D_1= \omega \cap AD_2$. Since $AD_2$ intersects $BC$ on the touching point of $A$-excircle, and $D_1P \perp D_1D_2$, we have that $DP=DD_1$, so $DD_1$ touches $\omega$. Furthermore, $D_1QD_2R$ is a harmonic quadrilateral, so $\frac{D_1Q}{D_1R}=\frac{D_2Q}{D_2R}=\frac{sin \angle IPQ}{sin \angle IPR}=\frac{sin \frac{C}{2}}{sin \frac{B}{2}} (\star)$ By Ratio Lemma, we know that $$\frac{XF}{XE}=\frac{sin \angle XDF}{sin \angle XDE} \frac{DF}{DE}=\frac{sin \angle XDF}{sin \angle XDE}\frac{b}{c}$$Since $DF \parallel AC$, $\angle XDF= \angle D_1DF= 180º- \angle D_1IQ$, so $sin \angle XDF= sin \angle D_1IQ= \frac{D_1Q}{2r}$, where $r$ is the inradius of $ABC$. Similarly, $sin \angle XDE= \frac{D_1R}{2r} \implies \frac{sin \angle XDF}{sin \angle XDE}= \frac{D_1Q}{D_1R}$, which is equals to $\frac{sin \frac{C}{2}}{sin \frac{B}{2}}$, from $(\star)$. Therefore, $\frac{XF}{XE}=\frac{sin \frac{C}{2}}{sin \frac{B}{2}}\frac{b}{c}$. Analogously, $\frac{YE}{YD}=\frac{sin \frac{B}{2}}{sin \frac{A}{2}}\frac{a}{b}$ and $\frac{ZD}{ZF}=\frac{sin \frac{A}{2}}{sin \frac{C}{2}}\frac{c}{a}$. Multiplying everything, we have that $$\frac{XF}{XE}.\frac{YE}{YD}.\frac{ZD}{ZF}=1$$so we are done by Menelaus' Theorem on triangle $DEF$ WRT line $XYZ$. $\blacksquare$

orz

a proof of the generalisation of the problem is stated in my post 2 posts underneath this one: https://artofproblemsolving.com/community/c6h1397197p30861050

InterLoop wrote: orz

Is the following lemma known by carnot's theorem? Lemma:- A circle meets the interior of $BC,CA,AB$ at $A_1,A_2;B_1,B_2;C_1,C_2$ respectively. If $AA_1,BB_1,CC_1$ concur then so does $AA_2,BB_2,CC_2$ concur

GeoKing wrote: Is the following lemma known by carnot's theorem? Lemma:- A circle meets the interior of $BC,CA,AB$ at $A_1,A_2;B_1,B_2;C_1,C_2$ respectively. If $AA_1,BB_1,CC_1$ concur then so does $AA_2,BB_2,CC_2$ concur this is a special case! Carnot's theorem states that even if any six points as defined above lie on a conic, the relation holds anyways, here is a proof of a generalisation of the problem (instead of the medial triangle, consider any cevian triangle).

Change the tangency points to $D,E,F$ and midpoints to $M,N,P$. Let $D',E',F'$ be the antipodes of $D,E,F$ on the incircle. Let $AD',BE',CF'$ meet the incircle at $T_D,T_E,T_F$ for second time. Note that $MT_D,NT_E,PT_F$ are tangent to incircle by homothety. In order to use menelaus, \[\Pi{(\frac{XN}{XP})}=\Pi{(\frac{\frac{\sin NMT_D.MN}{\sin MXN}}{\frac{\sin PMT_D.MP}{\sin MXN}})}=\Pi{(\frac{\sin MNT_D}{\sin PMT_D})}.\Pi{(\frac{MN}{MP})}=1\]We observe that $\Pi{(\frac{MN}{MP})}=1$. \[\Pi{(\frac{\sin MNT_D}{\sin PMT_D})}=\Pi{(\frac{\sin FIT_D}{\sin EIT_D})}=\Pi{(\frac{T_DF}{T_DE})}=\Pi{(\frac{\sin FD'A}{\sin AD'E})}=\Pi{(\frac{AF.\frac{\sin \frac{B}{2}}{AD'}}{AE.\frac{\sin \frac{C}{2}}{AD'}})}=1\]As desired.$\blacksquare$