In the previous lesson we considered cevians of a triangle and when they are concurrent. In this lesson we consider when three points, each on a different side of a triangle, are collinear. Like Ceva’s theorem, Menelaus' theorem shows that a geometrical condition, collinearity, is equivalent to an arithmetical condition for ratios. The two theorems are very similar, and in a subsequent lesson we will show that they are in fact equivalent. So, logically speaking, we don’t need to give a separate proof for each theorem. We do so nevertheless to exercise our understanding of this geometry.

In contrast to the proof of Ceva’s theorem, we shall first analyze the situation of points and lines and arrive at a place where an "if-and-only-if" proof can be given. Compared to our proof of Ceva’s theorem, this proof is more transparent because both parts are shorter, and have less redundancy. We need a definition and a restatement of the Main Collinearity Theorem that looks decidely odd, until you see how this "work" in our analysis.

Definition.

A line that crosses two other lines, but not at their intersection, is called a transversal of the two lines. For parallel lines, every line that crosses both is a transversal for them.

Comment.

All we have done to the wording of the original MCL is change some + signs to $-$ signs, and we left off the last condition. No new proof is needed. We take the original wording, but replace $+b$ by $-(-b)$. This gives us the pair $a, (-b)$ that plays the role of the pair $a, b$ in the reformulated wording. Remember, both wordings claim that there exists a number that plays a certain role. If we now replace $(-b)$ by $b$ it does not change what the statement says, just how we label numbers.

We now have three formulations of the MCL. The first is with the + signs. The second with the $-$ signs but, since we’re still talking about the same pair $a,b$ we must write $(-b)$. The letter $b$ in the first and second formulation refers to the same real number. But in the the third formulation the letter $b$ now refers to the negative of the number so labelled in formulations one and two.

Please remember that one reason that Euclidean Geometry continued to be taught for over 2000 years is that it trains the mind to think logically.

We left off the final collinearity condition and we reword with an entirely new set of letters to make it work for either version. Three points are collinear, $(PQR)$ if and only if there exists three non-zero numbers adding to zero, $p,q,r \in \mathbb{R}, 0 \ne pqr, 0 = p+q+r$, for which the linear combination $pP + qQ + rR = O$ is the origin.

Drawing Exercise.

You may use KSEG or a ruler and complass for this construction.

For $P \ne Q$ the displacement vectors $P - O$ and $Q - O$ are independent. It is customary to use the same letters since $P = P - O$. To find the $x,y$ for which $R = xP + yQ$ construct a line $\ell$ through $R$ and parallel to $(OP)$. Let $xQ$ be the point $\ell (OQ)$. Because we can write $xQ = (1-x)O + x Q \and (1-x)+x=1$, we know that every point of $(OQ)$ has such an $x$.

Similarly, let $yP = k(OP)$ where $(kR) \and k\ ||\ (OQ)$. Vector addition then reveals the resolution of $R$ as a linear combination of $P, Q$.

Question 1.

From the above it would seem that for $P \ne Q$ and every $R$ we can find $p,q,r$ for which $pP + qQ +rR = O$. But not just any three points are collinear. What gives?

Did you forget the second condition, $p + q + r =0$ ? In terms of $R = xP + yQ$ , we can take $p=x, q=y, r=-1$. Now we see that $0 = p+q+r = x+y - 1$ iff $x = 1-y$. In which case, $R = xP + (1-x)Q)$, hence $(PQR)$.

We are ready to proceed by applying MCLM several times.

 $(A'BC) \iff (\exists b,c \in \mathbb{R}, b \ne c, bc \ne 0) ( \frac{A' - B}{A' - C} = \frac{c}{b} )$ (1) $(B'CA) \and 0\ne c \in \mathbb{R} \iff (\exists! \ a \in \mathbb{R}, 0 \ne a \ne c))( \frac{B' - C}{B' - A} = \frac{a}{c} )$ (2)

Note the difference. In the first case the $b,c$ are not unique, as long as their ratio is the same. In the second case the denominator $c$ is already given, so the numerator $a$ is uniquely determined.

Next rewrite both fractional equations according to the first line of the MCLM, clear denominators, and add:

 $(b-c) A' = bB - cC$ $(c-a) B' = cC - aA$ $(b-c)A' + (c-a) B' = bB - aA$ (3)

If $b=a$, call it $t:=a=b$, then equation (3) becomes $(t-c) (A'-B') = t(B-A)$ with $t\ne 0 \ne t-c$, so $(A'B')\ ||\ (AB)$. Otherwise we can divide (3) by $b-a$ to obtain a point on both $(A'B')$ and $(AB)$.

 $(A'B')(AB) = \frac{b-c}{b-a}A' + \frac{c-a}{b-a}B' = \frac{bB - aA}{b-a}$ (4)

Now if $(C'AB) \and (A'B'C')$ then $C'$ is exactly the point in (4). From the MCLM we have that $\frac{C'-A}{C'-B} = \frac{b}{a}$. Multiplying:

 $\frac{A'-B}{A'-C} \frac{B'-C}{B'-A} \frac{C'-A}{C'-B} = \frac{c}{b}\frac{a}{c}\frac{b}{a} = 1$

Conversely, suppose we assume that

 $\frac{A'-B}{A'-C} \frac{B'-C}{B'-A} \frac{C'-A}{C'-B} = 1$

Then, substituting from (1) and (2), we obtain

 $\frac{c}{b} \frac{a}{c} \frac{C'-A}{C'-B} = 1$ $\implies \frac{C'-A}{C'-B} = \frac{b}{a}$ $\implies C' = \frac{bB-aA}{b-a}$ $\implies C' = \frac{aA-bB}{a-b}$ $\implies (a-b)C' = aA-bB$ (5)

Adding (3) and (5) yields the condition for collinearity:

 $(b-c)A' + (c-a) B' + (a-b)C' = O$ (5)

since none of the coefficients in (5) are zero while their sum is zero, $(b-c)(c-a)(a-b) \ne 0 = (b-c)+(c-a)+(a-b)$.

Thus we have proved this theorem:

Comment.

The structure of the two proofs, Ceva and Menelaus, can be best understood in terms of symbolic logic. In both cases we have the same hypothesis $\Sigma \equiv( \triangle ABC \and (A'BC), (B'CA), (C'AB))$ and a conclusion of the form $\Phi \iff \Psi$ .

To prove that $\Sigma \implies ( \Phi \iff \Psi )$ we need to prove that $\Sigma \Rightarrow ( \Phi \implies \Psi )$ and $\Sigma \Rightarrow ( \Psi \implies \Phi )$.

Question 2.

Why is $(\Sigma \implies ( \Phi \implies \Psi )) \equiv ((\Sigma \and \Phi) \implies \Psi)$ ?

Recall that a material implication $(\Sigma \implies \Xi)$ is equivalent to $(\not \Sigma \or \Xi)$. Just check their truth values and see they are the same. Now consider
 $\Sigma \implies ( \Phi \implies \Psi )$ $\equiv \not \Sigma \or ( \not \Psi \or \Phi )$ $\equiv ( \not \Sigma \or \not \Psi) \or \Phi )$ $\equiv \not( \Sigma \and \Psi) \or \Phi )$ $\equiv ( \Sigma \and \Psi) \implies \Phi )$