# 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:

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.)

• And what happens when we get all those -1s? – idliketodothis Feb 22 '17 at 21:23
• @idliketodothis It follows that $A_2,B_2,C_2$ are collinear by (the converse of) Menelaus' theorem. – dxiv Feb 22 '17 at 23:21
• But what I got for point B2 is B2C/B2B = -BA1/A1C; for point C1: AC2/CC2 = -B1A/CB1; All in all, I have C1B/AC1 * BA1/A1C * B1A/CB1 = -1. What am I doing wrong? – idliketodothis Feb 23 '17 at 6:19
• @idliketodothis Don't know how you somehow reverted to $(1)\,$. What you get in fact is $BA_2/A_2A \cdot AC_2/C_2C \cdot CB_2 / B_2B = -1\,$, which proves that $A_2,B_2,C_2$ are collinear. As a side note, you did not choose the best notation to preserve the symmetry in $A,B,C\,$, and it would have been easier to follow had you defined $AB \cap A_1B_1=C'$ instead of $A_2\,$. – dxiv Feb 23 '17 at 7:45