Given $3$ different points on line $p$ : $A,B,C$, and $3$ different points on line $q$ : $A',B',C'$.
In addition: $A'B \parallel AB'$, and $A'C \parallel AC'$.
The lines $p$ and $q$ are intersecting at point $M$ which is different from the points that mentioned.
Need to prove that $BC' \parallel B'C$.
MY ATTEMPT: I tried to use triangles similarity to find another similarity for triangles $B'A'C$ and $ABC'$ but didn't succeeded.
Thank you.



Consider the triangles $MA'B$ and $MB'A$, they are similar, so $$ \frac{MB'}{MA'}=\frac{MA}{MB}. $$ Consider now the triangles $MC'A$ and $MA'C$, they are similar, so $$ \frac{MC'}{MA'}=\frac{MA}{MC}. $$ Now, putting together these two equations we get $$ \frac{MC'}{MB'}=\frac{MB}{MC} $$ this means that the triangles $MB'C$ and $MC'B$ are similar, therefore $B'C||C'B$.