Ceva's Theorem

Ceva's Theorem

Given a $\triangle ABC$, as shown in graph. If Cevians $AD$, $BE$, $CF$ all pass through pint $O$, then according to Italian mathematician Giovanni Ceva, we have Ceva's Theorem says:

$$\frac{\overline{AD}}{\overline{DB}}\cdot \frac{\overline{BE}}{\overline{EC}} \cdot \frac{\overline{CF}}{\overline{FA}}=1$$

Where $\overline{AD}$ represents the length of segment $AD$

Derivation

Notice that the segment ${AD}$ and segment ${DB}$ share a same base $AB$.

$$\begin{align*} \implies \frac{\overline{AD}}{\overline{DB}}&=\frac{[\triangle AOD]}{[\triangle BOD]}=\frac{[\triangle ACD]}{[\triangle BCD]}\\ \therefore \frac{\overline{AD}}{\overline{DB}}&=\frac{[\triangle ACD]-[\triangle AOD]}{[\triangle BCD]-[\triangle BOD]}= \frac{[\triangle ACO]}{[\triangle BCO]} \end{align*}$$

Similarly, we may also et:

$$\frac{\overline{BE}}{\overline{EC}}=\frac{[\triangle BAO]}{[\triangle CAO]}, \ \frac{\overline{CF}}{\overline{FA}}=\frac{[\triangle CBO]}{[\triangle ABO]}$$

Hence:

$$\frac{\overline{AD}}{\overline{DB}}\cdot \frac{\overline{BE}}{\overline{EC}} \cdot \frac{\overline{CF}}{\overline{FA}}= \frac{[\triangle ACO]}{[\triangle BCO]}\cdot \frac{[\triangle ABO]}{[\triangle ACO]}\cdot \frac{[\triangle BCO]}{[\triangle ABO]}=1 \tag*{$\blacksquare$}$$

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