Menelaus's Theorem

Menelaus's Theorem

Menelaus's theorem is named for Greek Mathematician Menelaus of Alexandria that said: given a line and a $\triangle ABC$, if the line cross through the sides of $\triangle ABC$ $AB$, $AC$, and $BC$ or their extension at point $L$, $M$, and $N$ as shown in graph, then:

$$\frac{\overline{AM}}{\overline{MB}}\cdot \frac{\overline{BL}}{\overline{LC}} \cdot \frac{\overline{CN}}{\overline{NA}}=1$$

where $\overline{AM}$ represents the length of segment $AM$

Derivation

Derivation 1: Areas

Connect $AL$, as shown on the left.

Since there are some triangles share the same base and altitude:

$$\frac{\overline{AM}}{\overline{MB}}=\frac{[\triangle ALM]}{[\triangle BLM]}, \frac{\overline{BL}}{\overline{LC}}=\frac{[\triangle BLM]}{[\triangle CLM]}, \frac{\overline{CN}}{\overline{NA}}=\frac{[\triangle CLM]}{[\triangle ALM]}$$

Hence:

$$\frac{\overline{AM}}{\overline{MB}}\cdot \frac{\overline{BL}}{\overline{LC}} \cdot \frac{\overline{CN}}{\overline{NA}}=\frac{[\triangle ALM]}{[\triangle BLM]}\cdot \frac{[\triangle BLM]}{[\triangle CLM]} \cdot \frac{[\triangle CLM]}{[\triangle ALM]} =1 \tag*{$\blacksquare$}$$

Note: $[\triangle AOD]$ represents the area of $\triangle AOD$

---

Derivation 2: Trigonometry

Let $\angle AMN=\angle BML=\alpha$, $\angle ANM=\beta$, $\angle BLM=\gamma$.

According to the Law of Sines, in $\triangle AMN$:

$$\frac{\overline{NA}}{\sin{\alpha}}=\frac{{\overline{AM}}}{\sin{\beta}} \implies \frac{\overline{AM}}{\overline{NA}}= \frac{\sin{\beta}}{\sin{\alpha}}$$

Similarly, in $\triangle{BLM}$ and $\triangle{CLM}$, we get:

$$\frac{\overline{BL}}{\overline{MB}}= \frac{\sin{\alpha}}{\sin{\gamma}},\ \frac{\overline{CN}}{\overline{LC}}= \frac{\sin{\gamma}}{\sin{(180\degree-\beta)}}= \frac{\sin{\gamma}}{\sin{\beta}}$$

Hence:

$$\frac{\overline{AM}}{\overline{MB}}\cdot \frac{\overline{BL}}{\overline{LC}} \cdot \frac{\overline{CN}}{\overline{NA}}= \frac{\overline{AM}}{\overline{NA}}\cdot \frac{\overline{BL}}{\overline{MB}}\cdot \frac{\overline{CN}}{\overline{LC}}= \frac{\sin{\beta}}{\sin{\alpha}}\cdot \frac{\sin{\alpha}}{\sin{\gamma}} \cdot \frac{\sin{\gamma}}{\sin{\beta}}=1 \tag*{$\blacksquare$}$$

Examples