The problem you stated was #22. The actual problem 20 is: Quote: Point $D$ lies on side $AB$ of triangle $ABC.$ Let $\omega_1$ and $\Omega_1,\ \omega_2$ and $\Omega_2$ be the incirles and the excircles (touching segment $AB$) of triangles $ACD$ and $BCD.$ Prove that the common external tangents to $\omega_1$ and $\omega_2,$ $\Omega_1$ and $\Omega_2$ meet on $AB.$ [Mod: fixed!] Solution Denote the centres of $\omega_1, \omega_2, \Omega_1, \Omega_2$ as $I_1, I_2, I_3, I_4$ respectively. Note that the direct common tangents to $\omega_1, \omega_2$ intersect at a point collinear with $I_1,I_2,$ and same for $\Omega_1, \Omega_2.$ Since the intersection of the common tangents to $\omega_1, \omega_2$ is $AB\cap I_1I_2,$ and that for $\Omega_1, \Omega_2$ is $AB\cap I_3I_4;$ thus, our original problem can be reduced to showing that $AB,I_1I_2,I_3I_4$ concur. Note that $AI_1\cap BI_2$ is the incentre of $ABC;$ $AI_3\cap BI_4$ the $C$-excentre, and $I_3I_1\cap I_2I_4=C.$ Thus, the three points $AI_1\cap BI_2, AI_3\cap BI_4, I_3I_1\cap I_2I_4$ all lie on the $C$-bisector of $ABC.$ Thus, the triangles $AI_1I_3$ and $BI_2I_4$ are perspective by using Desargues theorem. Therefore, $AB, I_1I_2$ and $I_3I_4$ are concurrent. We are done.$\Box$

Let the internal and external centres of simultude of $\omega_1,\omega_2$ be $J$ and $K$, and the internal and external centres of simultude of $\Omega_1,\Omega_2$ be $L$ and $M$. Clearly $K, M \in AB$ and $C,J, L$ are collinear. Then $C(I_1,I_2;J,K)=C(I_3,I_4;L,M)=-1 \Longrightarrow K=M$ and the result follows.

My solution: Let $ T \in AB $ be the exsimilicenter of $ (\omega_1) \sim (\omega_2) $ . Let $ S $ be the exsimilicenter of $ (\Omega_1) \sim (\omega_2) $ . From D'Alembert theorem (for $ (\omega_1), (\omega_2), (\Omega_1) $ ) we get $ T \in CS $ . From D'Alembert theorem (for $ (\omega_2), (\Omega_1), (\Omega_2) $ ) we get $ T $ is the exsimilicenter of $ (\Omega_1) \sim (\Omega_2) $ . ie. the common external tangent of $ \{ (\omega_1), (\omega_2) \} $ and $ \{ (\Omega_1), (\Omega_2) \} $ are concurrent at $ T $ Q.E.D

Application to Geometry of Triangle (no need of the excenters Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P. Denote: Iab, Iac = the incenters of AA'B, AA'C, resp. A* = BC /\ IabIac. Similarly B*, C*. Which is the locus of P such that A*,B*,C* are collinear? APH

Here's my solution.

nsun48 wrote: Point $D$ lies on side $AB$ of triangle $ABC$. Let $\omega_1$ and $\Omega_1,\omega_2$ and $\Omega_2$ be the incircles and the excircles (touching segment $AB$) of triangles $ACD$ and $BCD.$ Prove that the common external tangents to $\omega_1$ and $\omega_2,$ $\Omega_1$ and $\Omega_2$ meet on $AB$. Let $I_1$, $I_2$, $E_1$ and $E_2$ denote the centers of $\omega_1$, $\omega_{2}$, $\Omega_1$ and $\Omega_2$ respectively. We know that for a pair of circles the common external tangents and the line joining their centers concur at a point, hence it suffices to show that $I_1I_2$, $AB$ and $E_1E_2$ are concurrent. Note that lines $AI_1$ and $BI_2$ concur at the incenter $(I)$ of $\triangle ABC$ $\implies I = AI_1 \cap BI_2$, and similarly, (the $C$-excenter)$ I_C = AE_1 \cap BE_2$. Next observe that $C, I_1, E_1$ and $C, I_2, E_2$ are collinear (by Fact5)$\implies I_1E_1 \cap I_2E_2=C$. Now we know that $C, I, I_C$ are collinear (again by Fact5), that is the point of intersections $I_C = AE_1 \cap BE_2$, $I = AI_1 \cap BI_2$ and $C=I_1E_1 \cap I_2E_2$ are collinear $\implies \triangle AI_1E_1$ and $BI_2E_2$ are perspective $\implies AB, I_1I_2$ and $E_1E_2$ are collinear. (by Desargues' theorem) $\quad \blacksquare$ Comment: Here also Fact5

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -14.844543093983653, xmax = 11.727496423492763, ymin = -5.2904326623236235, ymax = 11.914962925242275; /* image dimensions */ pen qqwwzz = rgb(0,0.4,0.6); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1); pen zzffff = rgb(0.6,1,1); pen wwccff = rgb(0.4,0.8,1); pen cczzff = rgb(0.8,0.6,1); /* draw figures */ draw((6,-1)--(1.2203061396318646,4.220306139631864), linewidth(0.8) + qqwwzz); draw(circle((1.297994792664679,7.748412115639051), 2.439813486998713), linewidth(0.8) + xdxdff); draw(circle((-6.230278041349824,4.384367495289647), 5.384367495289648), linewidth(0.8) + zzffff); draw(circle((3.2178857992303795,4.1763195717296195), 1.4436053237072248), linewidth(0.8) + xdxdff); draw(circle((1.1523497010570949,1.1341731242668103), 2.134173124266811), linewidth(0.8) + zzffff); draw((7.535809260498278,10.535809260498278)--(1.1523497010570949,1.1341731242668103), linewidth(0.8) + linetype("4 4")); draw((7.535809260498278,10.535809260498278)--(-6.230278041349824,4.384367495289647), linewidth(0.8) + linetype("4 4")); draw((6,-1)--(-8.77995987511004,10.539843958089863), linewidth(0.8)+dotted); draw((4,7)--(-4,-1), linewidth(0.8) + qqwwzz); draw((6,-1)--(-4,-1), linewidth(0.8) + qqwwzz); draw((6,-1)--(4,7), linewidth(0.8) + qqwwzz); draw((1.297994792664679,7.748412115639051)--(4,7), linewidth(0.8) + wwccff); draw((4,7)--(3.2178857992303795,4.1763195717296195), linewidth(0.8) + wwccff); draw((3.2178857992303795,4.1763195717296195)--(1.297994792664679,7.748412115639051), linewidth(0.8) + wwccff); draw((-6.230278041349824,4.384367495289647)--(-4,-1), linewidth(0.8) + cczzff); draw((-4,-1)--(1.1523497010570949,1.1341731242668103), linewidth(0.8) + cczzff); draw((1.1523497010570949,1.1341731242668103)--(-6.230278041349824,4.384367495289647), linewidth(0.8) + cczzff); draw((3.2178857992303795,4.1763195717296195)--(2.533748623874719,1.7063672931454548), linewidth(0.8)); draw((2.533748623874719,1.7063672931454548)--(1.1523497010570949,1.1341731242668103), linewidth(0.8)); draw((4,7)--(7.535809260498278,10.535809260498278), linewidth(0.8) + qqwwzz); /* dots and labels */ dot((4,7),linewidth(2pt) + dotstyle); label("$A$", (4.220895259805675,6.86627541692178), NE * labelscalefactor); dot((-4,-1),linewidth(2pt) + dotstyle); label("$B$", (-4.265549861088354,-1.520524555781675), NW * labelscalefactor); dot((6,-1),linewidth(2pt) + dotstyle); label("$C$", (6.163975649521138,-1.2548041606069118), NE * labelscalefactor); dot((1.2203061396318646,4.220306139631864),linewidth(2pt) + dotstyle); label("$D$", (0.8661752707242781,4.308716613364687), N * labelscalefactor); dot((1.297994792664679,7.748412115639051),linewidth(2pt) + dotstyle); label("$J_1$", (1.0986806165021967,8.111839769303481), NE * labelscalefactor); dot((-6.230278041349824,4.384367495289647),linewidth(2pt) + dotstyle); label("$J_2$", (-6.7068559917565,4.408361761555223), NW * labelscalefactor); dot((3.2178857992303795,4.1763195717296195),linewidth(2pt) + dotstyle); label("$I_1$", (3.4071265495829604,3.9101360206025424), NE * labelscalefactor); dot((1.1523497010570949,1.1341731242668103),linewidth(2pt) + dotstyle); label("$I_2$", (1.0488580424069285,0.7048837538069647), S * labelscalefactor); dot((2.533748623874719,1.7063672931454548),linewidth(2pt) + dotstyle); label("$I$", (2.526927740566554,1.1865019700612225), SW * labelscalefactor); dot((-8.77995987511004,10.539843958089863),linewidth(2pt) + dotstyle); label("$I_C$", (-8.915656776646728,10.137957782511048), NW * labelscalefactor); dot((7.535809260498278,10.535809260498278),linewidth(2pt) + dotstyle); label("$X$", (7.741690495871301,10.420285702384234), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]Let $I_1,I_2,J_1,$ and $J_2$ be the centers of $\omega_1,\omega_2,\Omega_1,$ and $\Omega_2,$ respectively. Notice that $I=\overline{AI_1}\cap\overline{BI_2},$ $I_C=\overline{AJ_1}\cap\overline{BJ_2},$ and $C=\overline{I_1J_1}\cap\overline{I_2J_2}$ are the incenter, $C$-excenter, and $C$-vertex of $\triangle ABC,$ respectively; hence, they are collinear. Then, by Desargues' Theorem, $\triangle AI_1J_1$ and $\triangle BI_2J_2$ are perspective and therefore $\overline{AB},\overline{I_1I_2},$ and $\overline{J_1J_2}$ concur. $\square$