This is not homework.
In triangle ABC,$C_1$ is the midpoint of AB. P is a point on $CC_1$. AP intersect BC at $A_1$ and BP intersect AC at $B_1$. Prove that $A_1B_1$ is parallel to AB.
My attempt: Extend $CC_1$ to C' so $CC_1$=$C_1C'$. Now ABC and ABC' are symmetrical. P' is on CC' and $PC_1$=$P'C_1$. Construct $A_1'$ and $B_1'$ similarly. ABPP' and CBC'A are both parallelograms. What should I do now?
Sorry I don't know how to put a diagram on.
EDIT:This is the diagram:
By Ceva's Theorem, $$\frac{AC_1}{C_1B}\cdot\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}=1$$
As $AC_1=C_1B$,
$$\frac{BA_1}{A_1C}=\frac{B_1A}{CB_1}$$
So, we have
$$\frac{BC}{A_1C}=\frac{AC}{B_1C}$$
$\triangle ABC\sim \triangle B_1A_1C$ and hence $A_1B_1\parallel BA$.