##### This post is devoted to Menelaus’s Theorem, named for the astronomer Menelaus of Alexandria. It is a very imporant theorem in geometry, relating the ratios of the segments obtained by “cutting” the sides of a triangle (or their extensions) with a straight line. Once you become familiar with it, you’ll find that it is very useful in solving efficienlty a wide range of problems that otherwise would require coordinate geometry and trigonometry.

Theorem. Consider a triangle $$\triangle ABC$$ and a line not passing through the triangle vertices. Let $$D$$, $$E$$ and $$F$$ its intersections with (the extension) of $$BC$$, $$AC$$, and $$AB$$ respectively, then

$$\frac{\overline{AF}}{\overline{FB}}\cdot \frac{\overline{BD}}{\overline{DC}}\cdot \frac{\overline{CE}}{\overline{EA}} = 1.\tag{1}\label{eq803:1}$$

The complete version of the Theorem, actually, states that the given ratio is a necessary and sufficient condition for the collinearity of points $$D$$, $$E$$, and $$F$$. Its demonstration is straightforward, and it is based on similarities among auxiliary triangles.

Let us practice the use of this Theorem with a first, simple, exercise.

From vertex $$A$$ of triangle $$\triangle ABC$$ consider a line that intersects the opposite side $$BC$$ in $$D$$, and from vertex $$B$$ a line that intersects $$AC$$ in $$E$$. Let $$P$$ the interesections between these lines. Suppose the area of $$\triangle ABP$$ is equal to $$6$$, the area of $$\triangle AEP$$ is equal to $$3$$, and the area of $$\triangle BDP$$ is $$4$$. Determine the area of the quadrilateral $$CDPE$$.

• Let $$x$$ be the unknown area of $$CDEP$$.
• In order to determine it, we will apply Menelaus’s Theorem to triangle $$\triangle BEC$$ cut by $$AD$$. So we have
$$\frac{\overline{EP}}{\overline{PB}}\cdot\frac{\overline{BD}}{\overline{DC}}\cdot\frac{\overline{AC}}{\overline{AE}}=1.\tag{2}\label{eq803:2}$$
• Our first aim is therefore writing LHS of \eqref{eq803:2} as a function of $$x$$.
• Observe that the ratios of the areas of $$\triangle AEP$$ and $$\triangle ABP$$ is equal to $$\frac{\overline{PE}}{\overline{PB}}$$. Why?
• Apply the same idea to the ratio between the areas of $$\triangle ABD$$ and $$\triangle ADC$$, which are respetively equal to $$10$$ and $$3+x$$, and thus obtain $$\frac{\overline{BD}}{\overline{DC}} = \frac{10}{3+x}$$.
• Again use the same approach on triangles $$\triangle ABC$$ and $$\triangle ABE$$ to express the ratio $$\frac{\overline{AC}}{\overline{AE}}$$ as a function of $$x$$.
• Use the previous results in \eqref{eq803:2} and solve the equation. You will find $$x = \frac{19}{2}$$.

We are ready now to approach a more structured exercise, that will better emphasize the efficiency of the method.

Let $$\triangle ABC$$ be an equilater triangle, and $$P$$ a point on the circumscribed circle. Let also $$D$$, $$E$$, and $$F$$ be the intersections of $$PA$$ with line $$BC$$, $$BP$$ with line $$AC$$ and $$PC$$ with line $$AB$$, respectively. Prove that the area of $$\triangle DEF$$ is twice the area of $$\triangle ABC$$.

Without loss of generality suppose $$P$$ lies in the half-plane determined by $$BC$$ and not containing $$A$$. We can also assume $$\overline{AB} = 1$$. In following the notation $$[\cdots]$$ will be used to indentify the area of the polygon whose vertices are listed in brackets.

• Observe that we can completely forget about the circle, since $$P$$ belonging to it is just equivalent to stating that $$\angle CPB = 120°$$. Why?
• Let us call $$\overline{BF} = x$$. Note that, once $$x$$ is fixed, the entire Figure is defined, so that our aim can be restated as computing $$[DEF]$$ as a function of $$x$$, and thus show that the result is eventually independent of $$x$$.
• We will first need some segment measures in $$\triangle BCF$$. The picture below, where $$CK$$, perpendicular to $$AB$$ has been drawn, shows all the details required.
• Prove that $$\triangle BCF$$ and $$\triangle BCP$$ are similar.
• Use this fact to show that $$\overline{CP}\cdot\overline{FC}= 1.\tag{3}\label{eq803:3}$$
• Observe that $$\triangle CKB$$ is half of an equilateral triangle. Compute then $$\overline{BK}$$ and $$\overline{KF}$$.
• Determine $$\overline{FC}$$ via Pythagorean Theorem. You should obtain $\overline{FC} = \sqrt{x^2+x+1}.$
• Use \eqref{eq803:3} to compute $\overline{CP}= \frac{1}{\sqrt{x^2+x+1}}$ and $\overline{FP} = \frac{x(x+1)}{\sqrt{x^2+x+1}}.$
• Getting back to the initial Figure, we can now repeatedly apply Menelaus’s Theorem to calculate all other segment lenghts we need. Start with $$\triangle ACF$$, cut by line $$BE$$. This will lead to $\frac{\overline{CE}}{\overline{AE}} = \frac{1}{x+1}.$ Since we have $$\overline{AE} – \overline{CE} = 1$$, we can get $$\overline{AE} = \frac{1}{x}$$ and $$\overline{CE} = \frac{x+1}{x}$$.
• Again our Theorem on $$\triangle BFC$$ and line $$AP$$, together with the information $$\overline{BD}+\overline{CD}=1$$, allows you to compute $$\overline{CD}$$ e $$\overline{BD}$$. Check your result with the following: $$\overline{CD} = \frac{1}{x+1}$$ and $$\overline{BD} = \frac{x}{x+1}$$.
• Switch now to $$\triangle ABC$$, intercepted by $$FG$$, and, using the Theorem and the relationship $$\overline{AG} + \overline{CG} = 1$$, determine $$\overline{AG} = \frac{x+1}{x+2}$$ and $$\overline{CG} = \frac{1}{x+2}$$.
• One last application of Menelaus’s Theorem to $$\triangle CGF$$ and line $$AP$$, will lead you to the ratio $\frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$
• Recalling the first exercise, we can go backward and use segment lengths to determine areas of triangles of fixed altitude and a known ratio between bases. Start with $$\triangle ACF$$ that shares with $$ABC$$ the altitude relative to $$AF$$. This implies $[ACF] = [ABC](1+x).$
• $$\triangle AFE$$ and $$\triangle ACF$$ have the altitude relative to $$AE$$ in common. Thus $[AFE] = [ACF] \cdot \frac{\overline{AE}}{\overline{AC}}.$ Check that this leads to $[AFE] = [ABC]\cdot \frac{(x+1)^2}{x}.$
• Do the same with $$\triangle GFE$$ and $$\triangle AFE$$ to obtain $$[GFE] = [ABC]\cdot\frac{2(x+1)^2}{x(x+2)}.\tag{4}\label{eq803:4}$$
• At last, observe that $$\frac{[DFE]}{[GDE]} = \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}\tag{5}\label{eq803:5}$$ and that $$[DFE] + [GDE] = [GFE]\tag{6}\label{eq803:6}.$$ Replace \eqref{eq803:4} in \eqref{eq803:6}. Determine from the latter equation $$[GDE]$$ and replace it in \eqref{eq803:5}. You will get an equation with $$[DEF]$$ as unknown, which, once solved, will give you $[DEF] = 2[ABC],$ as expected.