The key point is Lemma: Let $\mathcal E$ be an ellipse centred at $O$. Two points $A, B$ are chosen on $\mathcal E$. The tangents to $\mathcal E$ at $A, B$ meet at a point $X$. Then $XO$ bisects segment $AB$. In particular, $\triangle XAO {}$ and $\triangle XBO$ have the same area. Proof: Affine transform into a circle. Let's return to the main problem now. [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -368.5625926152787, xmax = 1121.2506999064312, ymin = -349.586248253878, ymax = 326.455330449013; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); draw((456.67750860203154,-57.17715508396357)--(468.20715990040446,-74.2186190737169)--(485.2486238901578,-62.68896777534398)--(473.71897259178485,-45.647503785590665)--cycle, linewidth(1.6) + zzttff); draw((556.200217195542,-130.84147716338492)--(544.6705658971691,-113.8000131736316)--(527.6291019074158,-125.3296644720045)--(539.1587532057887,-142.37112846175782)--cycle, linewidth(2) + zzttff); /* draw figures */ draw(circle((263.4137289679373,39.80426071872997), 227.0028624325864), linewidth(0.8)); draw((264.5024591913606,-187.1959908650256)--(473.71897259178485,-45.647503785590665), linewidth(0.8)); draw((264.5024591913606,-187.1959908650256)--(53.937822839803744,-47.660842633322), linewidth(0.8)); draw(circle((578.6642600297802,-45.144169073657835), 104.94649446873247), linewidth(0.8)); draw(shift((264.5024591913605,-187.1959908650256))*rotate(-179.7252022053201)*xscale(289.06743251828084)*yscale(140.54343410065127)*unitcircle, linewidth(0.8) + zzttff); draw((-10.571163986038627,-145.00772278881698)--(262.32499874451395,266.8045123024856), linewidth(0.8)); draw((262.32499874451395,266.8045123024856)--(539.1587532057887,-142.37112846175782), linewidth(0.8)); draw((53.937822839803744,-47.660842633322)--(683.6095474677754,-44.640834361725), linewidth(0.8)); draw((683.6095474677754,-44.640834361725)--(539.1587532057887,-142.37112846175782), linewidth(0.8)); draw((539.1587532057887,-142.37112846175782)--(263.82839771579427,-46.65417320945633), linewidth(0.8)); draw((262.32499874451395,266.8045123024856)--(264.5024591913606,-187.1959908650256), linewidth(0.8)); /* dots and labels */ dot((53.937822839803744,-47.660842633322),linewidth(2pt) + dotstyle); label("$A$", (34.92850744268443,-49.87771479735962), NE * labelscalefactor); dot((473.71897259178485,-45.647503785590665),linewidth(2pt) + dotstyle); label("$B$", (484.0063423629655,-39.20847897851916), NE * labelscalefactor); dot((262.32499874451395,266.8045123024856),linewidth(2pt) + dotstyle); label("$C$", (257.04259858036124,275.0490124127817), NE * labelscalefactor); dot((264.5024591913606,-187.1959908650256),linewidth(2pt) + dotstyle); label("$O$", (255.10273752239027,-209.9162520799665), NE * labelscalefactor); dot((263.82839771579427,-46.65417320945633),linewidth(2pt) + dotstyle); label("$D$", (246.37336276152087,-65.39660326112757), NE * labelscalefactor); dot((539.1587532057887,-142.37112846175782),linewidth(2pt) + dotstyle); label("$E$", (521.8336329933995,-161.41972563069172), NE * labelscalefactor); dot((683.6095474677754,-44.640834361725),linewidth(2pt) + dotstyle); label("$D'$", (692.5414060948455,-47.93785373938863), NE * labelscalefactor); dot((-10.571163986038627,-145.00772278881698),linewidth(2pt) + dotstyle); label("$F$", (-25.20718535441584,-146.87076769590925), NE * labelscalefactor); dot((401.4935754607915,-94.51265083560709),linewidth(2pt) + dotstyle); label("$M$", (389.9230810513731,-83.82528331185199), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Suppose the ellipse $\mathcal E$ touches the sides at $D, E, F$ respectively. Claim: $CE = CF$ and $D$ is the midpoint of $AB$. Proof: Using the Lemma on $E, F$ we obtain that $CO$ bisects $EF$; which implies that $CE = CF$, since $CO$ is the angle bisector of $\angle ECF$. Since $CE = CF$, it follows that $EF \parallel AB$. Consider line $CD$. It's pole wrt $\mathcal E$ is the intersection of lines $AB, EF$. Since this lies at infinity, it follows that $CD$ passes through the center of $\mathcal E$, that is, $C, D, O$ are collinear. This implies that $D$ is the midpoint of $AB$. Let $D'$ denote the reflection of $D$ over $B$ (note that this is a FIXED point, since $D$ is fixed (at the midpoint of $AB$) as well) Since $BO$ bisects $DE$, some simple homothety arguments centred at $D$ with factor $2$ easily show that we have $BE \parallel ED'$. In particular, $E$ varies on the circle with diameter $BD'$. It is easy to see that all points on this circle except the degenerate cases $B, D'$ are achievable as well. To conclude, the locus is the circle $(BD')$ where $D'$ is the point on ray $AB$ with $D'B/D'A = 1/3$.