The median $AB_1$ meets the side $A_1C_1$ of the medial triangle $A_1B_1C_1$ and $CP$ meets $AB$ in $Q$ show that $AB=3AQ$.
I tried to use Ceva's theorem but couldn't do that as according to Ceva's theorem if we consider P to be the meeting point then $$\frac{AQ}{QB}\frac{BB_1}{B_1C}\frac{CX}{XA}=1$$ I am taking the point of meeting of $BP$ in $AC$ as $X$. Now its clear $\frac{BB_1}{B_1C}=1$ but what to do for $\frac{CX}{XA}$.
I am new to using Menelaus theorem and Ceva's theorem , hence rectify me if I am wrong.
It is enough to use vectors. We have:
$$ A_1 = \frac{A+B}{2},\quad B_1=\frac{B+C}{2},\quad C_1=\frac{A+C}{2} $$ hence: $$ P = \frac{A_1+C_1}{2} = \frac{2A+B + C}{4}$$ and if $Q'$ is the point on $AB$ such that $\frac{AB}{AQ'}=3$, we have: $$ Q' = \frac{2A+B}{3} $$ and we just need to prove that $C,P,Q'$ are collinear, that is easy, since $4P=C+3Q'$.