ConstructionLet $X$ be the circumcenter of $\Delta ABC$, then construct $CX$ and name $C_1=CX\cap AB$. Then drop perpendiculars, $p_1,p_2$ from $C_1$ on $\overline{CA}$ and $\overline{CB}$ respectively and let $K_1=p_1\cap \overline{CA}$ and $K_2=p_2\cap \overline{CB}$. Then reflect $B$ over $K_2$ and name the new point as $A_1$(reflection can be made by constructing a circle with radius $BK_2$ and name the other intersection with $\overline{BC}$ as $A_1$). Do the same for $K_1$ and name the new point as $B_1$. Therefore, $A_1,B_1,C_1$ are the required points. Verification2nd condition: We can see that $\angle C_1B_1C=180-\angle C$ since $\angle C_1B_1A=\angle A$. Therefore we have $\angle CC_1B=180-(180-\angle A+90-\angle B)=\angle A+\angle B-90=90-\angle C$, similarly, we get $\angle CC_1A=90-C$, thus $CC_1$ bisects $\angle A_1C_1B_1$. 3rd Condition: Since $C_1A=C_1B_1$ and $C_1B=C_1A_1$, which implies $A_1C_1+C_1B_1=AB$. 1st Condition: We can see that \begin{align*} \frac{\sin(180-2A)}{B_1A}=\frac{\sin(A)}{C_1A}\implies B_1A=C_1A\cos(A)\\ \implies \frac{B_1A}{A_1B}=\frac{C_1A\cos(A)}{C_1B\cos(B)} \end{align*}In $\Delta CC_1A$, we can see that \begin{align*} C_1A=\frac{\sin(90-B)\cdot CC_1}{\sin(A)}=\frac{\cos(B)\cdot CC_1}{\sin(A)} \end{align*}thus we can say that \begin{align*} \frac{B_1A}{A_1B}=\frac{C_1A\cos(A)}{C_1B\cos(B)}=\frac{\sin(B)}{\sin(A)}=\frac{AC}{BC} \end{align*}Hence, we have $A_1B_1$ parallel to $AB$. Diagram[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 3.5, xmax = 8.5, ymin = 0.55, ymax = 5.5; /* image dimensions */ pen evevfz = rgb(0.8980392156862745,0.8980392156862745,0.9764705882352941); pen qqqqcc = rgb(0.,0.,0.8); pen ffevev = rgb(1.,0.8980392156862745,0.8980392156862745); pair A = (5.,5.), B = (4.,2.), C = (8.,2.), A_1 = (5.141302270851851,2.), B_1 = (5.8559767031388885,4.1440232968611115), C_1 = (4.566375937413335,3.699127812240006), X = (6.,3.); filldraw(A--B--C--cycle, evevfz, linewidth(0.8) + qqqqcc); filldraw(A_1--B_1--C_1--cycle, ffevev, linewidth(0.8) + red); /* draw figures */ draw(A--B, linewidth(0.8) + qqqqcc); draw(B--C, linewidth(0.8) + qqqqcc); draw(C--A, linewidth(0.8) + qqqqcc); draw(A_1--B_1, linewidth(0.8) + red); draw(B_1--C_1, linewidth(0.8) + red); draw(C_1--A_1, linewidth(0.8) + red); draw((xmin, -0.49067628120689016*xmin + 5.9397401758026085)--(xmax, -0.49067628120689016*xmax + 5.9397401758026085), linewidth(0.8)); /* line */ draw((xmin, 1.*xmin-0.8559767031388893)--(xmax, 1.*xmax-0.8559767031388893), linewidth(0.8)); /* line */ draw((4.570651135425925,ymin)--(4.570651135425925,ymax), linewidth(0.8)); /* line */ draw(circle(X, 2.23606797749979), linewidth(0.8)); /* dots and labels */ dot(A,linewidth(3.pt) + dotstyle); label("$A$", (5.052159914288718,5.087286353518244), NE * labelscalefactor); dot(B,linewidth(3.pt) + dotstyle); label("$B$", (3.7893098629776683,1.77416210125514), NE * labelscalefactor); dot(C,linewidth(3.pt) + dotstyle); label("$C$", (8.068142978008048,1.789019160682329), NE * labelscalefactor); dot(A_1,linewidth(3.pt) + dotstyle); label("$A_1$", (5.0373028548615295,1.7593050418279512), NE * labelscalefactor); dot(B_1,linewidth(3.pt) + dotstyle); label("$B_1$", (5.91386936106567,4.2404339661684824), NE * labelscalefactor); dot(C_1,linewidth(3.pt) + dotstyle); label("$C_1$", (4.235021645793333,3.6015804107993636), NE * labelscalefactor); dot(X,linewidth(3.pt) + dotstyle); label("$X$", (6.062439955337558,3.0964403902749438), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]