Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

enter image description here $$\frac{AJ}{JB} \cdot \frac{BI}{IC} \cdot \frac{CK}{EK} = 1$$

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.

share|cite|improve this question
The only proof I know of this uses barycentric coordinates, is valid over any field and thus makes no use of area. – Olivier Bégassat Nov 24 '12 at 12:48
up vote 5 down vote accepted

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 $ABC$.

share|cite|improve this answer

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

Ceva's Theorem

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

$\hspace{1.5cm}$enter image description here

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

Law of Sines

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

$\hspace{3.5cm}$enter image description here

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.