Prove product of 3 cross-ratios equals to 1 or -1

Question

Here is a problem from a Chinese textbook Advanced Geometry (4th edition, ISBN: 9787040537550):

A straight line intersects a triangle $P_{1}P_{2}P_{3}$'s 3 edges $P_{2}P_{3}$, $P_{3}P_{1}$ and $P_{1}P_{2}$ at point $Q_{1}$, $Q_{2}$ and $Q_{3}$. Take another 3 points $Q'_{1}$, $Q'_{2}$ and $Q'_{3}$ on $P_{2}P_{3}$, $P_{3}P_{1}$ and $P_{1}P_{2}$ respectively, such that $(P_{2},P_{3};Q'_{1},Q_{1})=k_1$, $(P_{3},P_{1};Q'_{2},Q_{2})=k_2$ and $(P_{1},P_{2};Q'_{3},Q_{3})=k_3$. Prove:

1. $Q'_{1}$, $Q'_{2}$ and $Q'_{3}$ are collinear if and only if $k_1\cdot k_2\cdot k_3=1$
2. $P_{1}Q'_{1}$, $P_{2}Q'_{2}$ and $P_{3}Q'_{3}$ are concurrent if and only if $k_1\cdot k_2\cdot k_3=-1$

I think statement 1 can be proved by using Menelaus's theorem twice, and statement 2 can be proved by combining Menelaus's theorem and Ceva's theorem. However, the next problem shows that my approach is not the answer the book wants:

Use above results to prove Menelaus's theorem and Ceva's theorem. (Hint: Place $Q_{1}$, $Q_{2}$ and $Q_{3}$ on an infinity line.)

Is there any other proofs of these 2 statement without Menelaus's theorem and Ceva's theorem? (Menelaus's theorem can be easily proved by sine law, so I think sine law may not be used in the proof.)

A star before problem 6 may indicate the problem is difficult.

Answer 1

I bought the guide book (ISBN: 9787040129472) for that textbook and got the hint to use a theorem stated in the textbook:

If four different collinear points are represented in homogeneous coordinates as $P_1(a)$, $P_2(b)$, $P_3(a+\lambda_1b)$ and $P_4(a+\lambda_2b)$, then $(P_1,P_2;P_3,P_4)=\lambda_1/\lambda_2$.

A simplified form of this theorem is stated here.

So we represent $Q_1$, $Q_2$, $Q_3$, $Q'_1$, $Q'_2$ and $Q'_3$ as:

$$\begin{cases} Q_1=P_2+\lambda_1P_3\\ Q_2=P_3+\lambda_2P_1\\ Q_3=P_1+\lambda_3P_2\\ Q'_1=P_2+\lambda'_1P_3\\ Q'_2=P_3+\lambda'_2P_1\\ Q'_3=P_1+\lambda'_3P_2\\ \end{cases}$$

such that $k_1=\lambda'_1/\lambda_1$, $k_2=\lambda'_2/\lambda_2$ and $k_3=\lambda'_3/\lambda_3$.

$Q_1$, $Q_2$ and $Q_3$ are collinear if and only if $Q_3$ can be represented as $Q_3=xQ_1+yQ_2$. Take $(P_1,P_2,P_3)$ as basis then we can solve x, y and $\lambda_3$, so $\lambda_3=-1/\lambda_1\lambda_2$.

So the first statement:

$Q'_1$, $Q'_2$ and $Q'_3$ are collinear if and only if $k_1\cdot k_2\cdot k_3=1$

This can be proved from analogous $\lambda'_3=-1/\lambda'_1\lambda'_2$.

Then the second statement:

$P_1Q'_1$, $P_2Q'_2$ and $P_3Q'_3$ are concurrent if and only if $k_1\cdot k_2\cdot k_3=-1$

Let point $O=P_1Q'_1\cap P_2Q'_2$ then $O=xP_1+Q'_1=yP_2+zQ'_2$. $P_1Q'_1$, $P_2Q'_2$ and $P_3Q'_3$ are concurrent at O if and only if $O=wP_3+vQ'_3$. Also take $(P_1,P_2,P_3)$ as basis then we can solve x, y, z, w, v and $\lambda'_3$, so $\lambda'_3=1/\lambda'_1\lambda'_2$, which proves the statement.