Ceva’s Theorem.

Three cevians of a triangle are concurrent if and only if Ceva’s equation holds. Or in other words, if `(A'BC)`, `(AB'C)`, and `(ABC')` then a geometrical condition is logically equivalent to an arithmetic condition:

`G=(A A')(B B')(C C')\ hArr\ \frac{A'-B}{A'-C}\frac{B'-C}{B'-A}\frac{C'-A}{C'-B} = -1`.

Proof of Ceva’s Theorem Part 1

First, will prove that the geometric condition on the left-hand side (LHS) of the equivalence implies the arithmetic condition on the RHS. From $ \triangle ABC $ we have a unique set of barycentric coordinates $ G=aA+bB+cC \and a+b+c=1 $.

To say that $ G(A'A) \and (A'A)$ is a cevian implies that $ (GA) $ is not parallel to $ (BC) $. So, by the final question in the previous lesson, we know that $ b+c \ne 0 $ and we can rewrite $ G = aA + (b+c)(\frac{bB+cC}{b+c}) $. Applying the Main Collinearity Lemma, we conclude that $ \frac{bB+cC}{b+c} = A' \and \frac{A'-B}{A'-C} = -\frac{c}{b} $

Now apply the same argument starting with vertex $ B $. That is, in the previous paragraph, replace A by B and B by C and C by A. The last line of your rewritten argument will end in $ \frac{B'-C}{B'-A} = -\frac{a}{c} $.

Cycling the letters once more yields9 $\frac{C'-A}{C'-B} = -\frac{b}{a} $. Multiplying the three equations together yields the arithmetic equation on the RHS in Ceva’s theorem.

Proof of Ceva’s Theorem Part 2

Here we prove the converse implication: if `(A'BC)`, `(AB'C)`, and `(ABC')` and

`\frac{A'-B}{A'-C}\frac{B'-C}{B'-A}\frac{C'-A}{C'-B} = -1`.

(1)

then, `(A A')(B B')(C C')`. We label this intersection point by `G`. In other words, given points `A'` on `(BC)`, `B'` on `(AC)`, and `C'` on `(AB)`, and assuming equation (1), our task will be to somehow find the point G, the intersection of the three cevians.

We’ll start by analyzing what we know: equation (1) is true. This means the product on the left side is a real number and that number is equal to `-1`. Hence we can conclude that each factor must be a real number and have nonzero numerator and denominator.

Question.

Why can’t we have any zeros?

Answer. [show]

Suppose one of the numerators is equal to 0. Then that fraction must be 0. Multiplying the other two fractions by 0 yields 0, not `-1`, contradicting the assumption!

On the other hand, if one of the denominators is 0, then that factor does not exist, and equation (1) wouldn’t make sense!

Now set

`\frac{A'-B}{A'-C}`

= `-\frac{c}{b}` for some `c,b\ne 0` and

(2)

`\frac{B'-C}{B'-A}`

= `-\frac{a}{c}` for some `a\ne 0`.

(3)

Let us see what happens when we plug this into equation (1).

`(-\frac{c}{b})(-\frac{a}{c})(\frac{C'-A}{C'-B}) = -1`

`(\frac{a}{b})(\frac{C'-A}{C'-B}) = -1`

`\frac{C'-A}{C'-B}=-\frac{b}{a}`.

(4)
Question.

Why can we make those particular choices of variables in (2) and (3)?

Answer. [show]

For equation (1), Apply the Main Collinearity Lemma by replacing $ (X,A,B) \leftarrow (A',B,C) $. For equation (2), however, we wish to re-use the number $ c $ which we have already found. The MCL guarantees some ratio, say $ -\frac{\alpha}{\beta} = \frac{B'-C}{B'-A} $. Arithmetically, we can force the denominator of this fraction to be $ c $ by making the numerator $ c \alpha \beta^{-1} $. Isn’t algebra wonderful!

It is essential that you get used to substituting letters in this business. Remember the roles played by various letters, not only in terms of the original values for the letters. You learned the ability to do this in high school algebra.

At this point, we have simply renamed the three factors in (1) in terms of some numbers `a,b`, and `c`. How does this help in actually finding `G`? Recall the Main Collinearity Lemma. Apply it to the variables in equation (2) means we can write

`A'=\frac{bB+cC}{b+c}`.

Now define

`G:=\frac{aA+bB+cC}{a+b+c}`.

Applying a little algebra, we can write

`G=\frac{aA+bB+cC}{a+b+c}=\frac{aA+ \frac{bB+cC}{b+c}(b+c)}{a+b+c}=\frac{aA+(b+c)A'}{a+b+c}`.

Which, by the MCL, means `(GA A')`. Now applying the same technique to (3), we get

`B'=\frac{aA+cC}{a+c}`.

and

`G=\frac{aA+bB+cC}{a+b+c}=\frac{bB+\frac{aA+cC}{a+c}(a+c)}{a+b+c}=\frac{bB+(a+c)B'}{a+b+c}`.

And so, `(GB B')`.

Question.

Repeat the above once more with (4) to get that `(GC C')`.

Answer. [show]

Again, by the MCL, (4) implies that

`C'=\frac{aA+bB}{a+b}`.

And so,

`G=\frac{aA+bB+cC}{a+b+c}=\frac{cC+\frac{aA+bB}{a+b}(a+b)}{a+b+c}=\frac{cC+(a+b)C'}{a+b+c}`,

which, by the MCL, means `(GC C')`.

To sum up, we have found some `G` such that `(GA A')`, `(GB B')`, and `(GC C')`; in other words, the three cevians of △`ABC` intersect at `G`.

Comment.

The proof of Ceva’s theorem as given in Tondeur’s textbook is not really that different from the above. But we have rearranged it by extracting the central idea, which we summarized in the MCL. Isolating a central idea of theme of a complicated proof is a good way to make the actual proof shorter, and therefore easier to follow. We shall do this many more times in this course.

You should study Tondeur’s proof as well. Perhaps you will like it better than the above. But it would be a bad idea to google Ceva’s theorem to find an easier proof. You will find many different proofs, but they will not be formulated from the vector viewpoint, which is essential for the remainder of this course.