Omid Hatami wrote: Let $ABC$, $l$ and $P$ be arbitrary triangle, line and point. $A',B',C'$ are reflections of $A,B,C$ in point $P$. $A''$ is a point on $B'C'$ such that $AA''\parallel l$. $B'',C''$ are defined similarly. Prove that $A'',B'',C''$ are collinear. http://www.cut-the-knot.org/Curriculum/Geometry/Zaslavsky.shtml (interchange triangles ABC and A'B'C'). darij

My solution: Let $A_3,B_3,C_3$ be points on $BC,CA,AB$ respectively,such that $A_2A_3 \| BC',B_2B_3 \| CA',C_2C_3 \| AB'$ Easy to see that $AB' = A'B = C_2C_3,BC' = B'C = A_2A_3,CA' = C'A = B_2B_3$.Then apply Thales's theorem and Menelaus's theorem we have:$A_2,B_2,C_2$ are collinear if and only if $A_3,B_3,C_3$ are collinear. Let $d_1,d_2,d_3$ be lines pass through $A,B,C$ and parallel to $A_2A_3,B_2B_3,C_2C_3$ ,respectively.Then $d_1,d_2,d_3$ are concurrent at Q (the proof is simple,just take $Q' = d_1 \cap d_2$ then $QB = CA'$ and $QB \| CA'$ so $QBA'C$ is parallelogram and from it we have $CQ \| BA' \| C_2C_3$. $QAA_2A_3,QBB_2B_3,QCC_2C_3$ are parallelograms so $QA_3 \| QB_3 \| QC_3 \| l$,then $Q,A_3,B_3,C_3$ are collinear and that complete the proof Image not found

Trivial by moving point.Move the point $A''$ on $B'C'$ then define $B''$and $C''$ such that $AA''//BB''//CC''$ $deg(A'')=deg(B'')=deg(C'')=1$ we want to show that $A''-B''-C''$ collinear so 4 case is enough.Take $A''$ $C''$,$B''$ ,$AA''//BB'$ and $\infty_{BC}$ .$\blacksquare$