On Ceva's Theorem?

Question

The famous Ceva's Theorem on a triangle $\Delta \text{ABC}$

$$\frac{AJ}{JB} \cdot \frac{BI}{IC} \cdot \frac{CK}{EK} = 1$$

is usually proven using the property that the area of a triangle of a given height is proportional to its base.

Is there any other proof of this theorem (using a different property)?

EDIT: I would like if someone can use the proof of Menelaus' Theorem.

Answer 1

The following proof is heavily influenced by my background in projective geometry. Use homogenous and choose the following affine basis, without loss of generality. You might also consider this as barycentric coordinates, the way Olivier Bégassat wrote in a comment.

\begin{align*} A &= \begin{pmatrix}1\\0\\0\end{pmatrix} & B &= \begin{pmatrix}0\\1\\0\end{pmatrix} & C &= \begin{pmatrix}0\\0\\1\end{pmatrix} \end{align*}

Based on these coordinates, you can specify the other three points as linear combinations of the corresponding triangle corners:

\begin{align*} I &= \begin{pmatrix}0\\\lambda_I\\\mu_I\end{pmatrix} & J &= \begin{pmatrix}\lambda_J\\\mu_J\\0\end{pmatrix} & K &= \begin{pmatrix}\mu_K\\0\\\lambda_K\end{pmatrix} \end{align*}

You can obtain the oriented length ratios from these parameters:

\begin{align*} \frac{AJ}{JB} &= \frac{\mu_J}{\lambda_J} & \frac{BI}{IC} &= \frac{\mu_I}{\lambda_I} & \frac{CK}{KA} &= \frac{\mu_K}{\lambda_K} \end{align*}

Now you can compute the connections of each of these points with the opposite triangle corner using a cross product:

\begin{align*} A\times I &= \begin{pmatrix}0\\-\mu_I\\\lambda_I\end{pmatrix} & B\times K &= \begin{pmatrix}\lambda_K\\0\\-\mu_K\end{pmatrix} & C\times J &= \begin{pmatrix}-\mu_J\\\lambda_J\\0\end{pmatrix} \\ \end{align*}

To check whether these three lines are concurrent, you compute their determinant. If that determinant becomes zero, the lines go through the same point.

\begin{align*} \begin{vmatrix} 0 & \lambda_K & -\mu_J \\ -\mu_I & 0 & \lambda_J \\ \lambda_I & -\mu_J & 0 \end{vmatrix} = \lambda_I\lambda_J\lambda_K - \mu_I\mu_J\mu_K &= 0 \\ \lambda_I\lambda_J\lambda_K &= \mu_I\mu_J\mu_K \\ 1 &= \frac{\lambda_I\lambda_J\lambda_K}{\mu_I\mu_J\mu_K} = \frac{AJ}{JB}\cdot\frac{BI}{IC}\cdot\frac{CK}{KA} \end{align*}

As the choice of coordinates was without loss of generality, the above equivalence (between the concurrence of the three lines and the product of oriented length ratios being one) will hold for any non-degenerate triangle $ABC$.

Answer 2

Here is a proof using the Law of Sines (assuming that is allowed).

Ceva's Theorem

The sines of supplementary angles are equal, so $$ \frac{\overline{CM}}{\overline{CE}}\sin(c)=\overbrace{\sin(\angle MEC)=\sin(\angle MEA)}^{\Large\text{supplementary sines}}=\frac{\overline{AM}}{\overline{AE}}\sin(a)\tag{1} $$ $$ \frac{\overline{AM}}{\overline{AF}}\sin(b)=\sin(\angle MFA)=\sin(\angle MFB)=\frac{\overline{BM}}{\overline{BF}}\sin(c)\tag{2} $$ $$ \frac{\overline{BM}}{\overline{BD}}\sin(a)=\sin(\angle MDB)=\sin(\angle MDC)=\frac{\overline{CM}}{\overline{CD}}\sin(b)\tag{3} $$

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Multiplication of $(1)$, $(2)$, and $(3)$, and cancelling yields Ceva's Theorem: $$ \frac{\overline{AE}\;\overline{BF}\;\overline{CD}}{\overline{CE}\;\overline{AF}\;\overline{BD}}=1\tag{4} $$

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Law of Sines

The Law of Sines can be proven using only the definition of sine:

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$$ a\sin(B)=h=b\sin(A)\tag{5} $$ Cancelling yields the Law of Sines: $$ \frac a{\sin(A)}=\frac b{\sin(B)}\tag{6} $$

Answer 3

I'm not sure what you meant while stating the usage of the proof of Menelaus Theorem, yet I can share with you another proof of Ceva Theorem by using Menelaus Theorem itself not the proof. Menelaus in $ABI$ with respect to $CJ$:

$\frac{CB}{CI} . \frac{OI}{OA}.\frac{AJ}{BJ} =1$

Menelaus in $AIC$ with respect to $BK$:

$\frac{BI}{BC}.\frac{KC}{KA}.\frac{OA}{OI} =1$

First I should state that we should consider all lengths as how we concern vectors, in this sence: $BC$=$-CB$ etc.

Then multiplying these two yields: $\frac{AJ}{BJ}.\frac{BI}{CI}.\frac{CK}{AK} =-1$

I've ended up with a minus sign, it is just because as I've told before if you compare the directions of each vector in both equations you will see that your result and mine are exactly the the same.

Answer 4

I’m sure there is a slick proof lurking in $\mathbb{C}$. This is not it.

We first prove the left implication. Place the origin $O$ at the concurrent point as figured.

Since ratios of lengths are invariant under dilations and rotations, WLOG let the line through $B$ and $K$ be the real axis and scale the triangle such that $B=1$.

As $I,J,K$ lie on the edges of $\triangle ABC$ $$ \begin{split} I&=A+t_1(1-A)\\J&=1+t_2(C-1)\\K&=C+t_3(A-C),\end{split}\tag{*}$$ for real $t_i$, from which it follows $$\tag{**}\frac{ A-I}{ I-B}\cdot \frac{ B-J}{ J-C}\cdot \frac{ C-K}{ K-A}=\frac{t_1t_2t_3}{(1-t_1)(1-t_2)(1-t_3)}.$$ Further, $$\begin{split} I&=r_1C\\J&=r_2A \\K&=r_3.\end{split}\tag{***}$$ If we equate $(*)$ and $(***)$ and solve for the real $r_i$, then we get three complex equations in $r_i,a_i,c_i, t_i$ with imaginary parts $0$. Solving these three imaginary parts for $t_i$ then gives $$\begin{split} t_1=\frac{a_1c_2-a_2c_1}{ a_1c_2-a_2c_1-c_2} &\implies \frac{1}{1-t_1}=\frac{c_2+a_2c_1-a_1c_2}{c_2}\\t_2=\frac{a_2}{a_1c_2+a_2-a_2c_1}&\implies \frac{1}{1-t_2}=\frac{a_1c_2+a_2-a_2c_1}{a_1c_2-a_2c_1}\\t_3=\frac{c_2}{c_2-a_2}&\implies \frac{1}{1-t_3}=\frac{a_2-c_2}{a_2}\end{split}$$ and plugging these into $(**)$ gives the desired result.

To prove the right implication we follow basically the same procedure, i.e. place the origin $O$ at the intersection of Ceva lines $CI$ and $AJ$, rotate and scale so that $B=1$ and then prove $K\in\mathbb{R}$ by assuming the Ceva product is $1$, hence the Ceva lines are concurrent. The same algebraic manipulations as before show $$\begin{split} \text{Im}(K)&=t_3a_2-c_2(1-t_3)\\&=t_3a_2+c_2\frac{t_1t_2t_3}{(1-t_1)(1-t_2)}\\&=t_3a_2 +c_2\frac{a_2c_1-a_1c_2}{c_2}\frac{a_2}{a_1c_2-a_2c_1}t_3\\&=0\end{split}$$ and we conclude the result. $\qquad\square$

Not nice, but maybe you can streamline the argument somewhere.