Prove that three points are collinear (Menelaus theorem?) Given a point P inside the triangle ABC. Lines AP, BP, CP intersect the sides BC, CA, AB in points $A_1, B_1, C_1$. The lines $A_1B_1$ and AB intersect in the point  $A_2$. The points $B_2,C_2$ are defined in a similar way. How can I prove that $A_2, B_2, C_2$ are collinear?
I think it has to do someting with the Menelaus theorem. But how can I use it here?. Another hard thing in this task is that it is almost impossible to draw a good figure, the best thing I could get is that:

 A: Hint: by Ceva's theorem for the cevians through $P\,$:
$$
\frac{BA_1}{A_1C}\cdot\frac{CB_1}{B_1A}\cdot\frac{AC_1}{C_1B} = 1
\tag{1}$$
By Menelaus' theorem for the transversal $A_1B_1\,$:
$$
\frac{BA_2}{A_2A}\cdot\frac{AB_1}{B_1C}\cdot\frac{CA_1}{A_1B} = -1
\tag{2}
$$
Multiplying the two $(1) \cdot (2)\,$:
$$
\require{cancel}
\cancel{\frac{BA_1}{A_1C}}\cdot\bcancel{\frac{CB_1}{B_1A}}\cdot\frac{AC_1}{C_1B} \;\cdot\; \frac{BA_2}{A_2A}\cdot\bcancel{\frac{AB_1}{B_1C}}\cdot\cancel{\frac{CA_1}{A_1B}} = -1 \;\;\iff\;\; \frac{BA_2}{A_2A} = -\,\frac{C_1B}{AC_1}
$$
Repeat for $B_2,C_2\,$, multiply together, and the RHS reduces to $-1$ using $(1)$ one more time. 
(It should be noted that the problem is equivalent to half the Desargues' theorem in the plane.)
A: A proof using projective geometry: Triangles $ABC$ and $A_1B_1C_1$ are perspective from the point $P$, so by Desargues' Theorem they are perspective from a line too, i.e. their pairs of corresponding sides intersect at $3$ points which are collinear. This implies $A_2,B_2,C_2$ are collinear.
