How can you prove Ceva's Theorem using vectors? Given the following layout, where $ \overrightarrow{CY} = \mathbf{a} $ and $ \overrightarrow{CX} = \mathbf{b} $:
                                             
Prove that $$ \frac{\overrightarrow{BX}}{\overrightarrow{XC}} \times \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \times \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} = 1$$
 A: Our proof is in three parts. Define $ \overrightarrow{CB} = \alpha\overrightarrow{CX}$, $ \overrightarrow{CA} = \beta\overrightarrow{CY}$ and $ \overrightarrow{AB} = \gamma\overrightarrow{AZ}$. Also, let $ \overrightarrow{AX} = \lambda\overrightarrow{AP}$, $ \overrightarrow{BY} = \mu\overrightarrow{BP}$ and $ \overrightarrow{CZ} = \nu\overrightarrow{CP}$.

1. $ \alpha, \beta, \overrightarrow{CP} $
$\overrightarrow{CP} = \overrightarrow{CX} + \overrightarrow{XP} = \mathbf{b} - \overrightarrow{PX} = \mathbf{b} - (\overrightarrow{AX} - \overrightarrow{AP}) = \mathbf{b} - (\overrightarrow{AX} - \frac{1}{\lambda}\overrightarrow{AX}) = \mathbf{b} - (\frac{\lambda - 1}{\lambda}(\overrightarrow{AC} + \overrightarrow{CX})) = $   $\mathbf{b} - \frac{\lambda - 1}{\lambda}\mathbf{b} + \frac{(\lambda - 1)\beta}{\lambda}\mathbf{a}$
By a similar argument, $ \overrightarrow{CP} = \overrightarrow{CY} + \overrightarrow{YP} = \mathbf{a} - \frac{\mu - 1}{\mu}\mathbf{a} + \frac{(\mu - 1)\beta}{\mu}\mathbf{b}$
So, $ \mathbf{b} - \frac{\lambda - 1}{\lambda}\mathbf{b} + \frac{(\lambda - 1)\beta}{\lambda}\mathbf{a} = \mathbf{a} - \frac{\mu - 1}{\mu}\mathbf{a} + \frac{(\mu - 1)\beta}{\mu}\mathbf{b}$
       $ (1 - \frac{\lambda - 1}{\lambda} - \frac{(\mu - 1)\alpha}{\mu})\mathbf{b} = (1 - \frac{\mu - 1}{\mu} - \frac{(\lambda - 1)\beta}{\lambda})\mathbf{a}$
From which it follows that $$ 1 - \frac{\lambda - 1}{\lambda} - \frac{(\mu - 1)\alpha}{\mu} = 0 \Rightarrow \beta = \frac{\lambda}{\lambda\mu - \mu} $$
and
$$ 1 - \frac{\mu - 1}{\mu} - \frac{(\lambda - 1)\beta}{\lambda} = 0 \Rightarrow \alpha = \frac{\mu}{\mu\lambda - \lambda} $$
We can substitute either one of these into either expression for $ \overrightarrow{CP}$ to get $$ \overrightarrow{CP} = \frac{1}{\mu}\mathbf{a} + \frac{1}{\lambda}\mathbf{b}$$

2. $ \gamma $
$ \overrightarrow{CZ} = \overrightarrow{CA} + \overrightarrow{AZ} = \beta \mathbf{a} + \frac{1}{\gamma}(\overrightarrow{AC} + \overrightarrow{CB}) = \beta \mathbf{a} - \frac{\beta}{\gamma}\mathbf{a} + \frac{\alpha}{\gamma}\mathbf{b}$
We substitute this result into $ \overrightarrow{CZ} = \nu\overrightarrow{CP} $ to get $$ \beta \mathbf{a} - \frac{\beta}{\gamma}\mathbf{a} + \frac{\alpha}{\gamma}\mathbf{b} =  \frac{\nu}{\mu}\mathbf{a} + \frac{\nu}{\lambda}\mathbf{b}$$
$$ (\beta - \frac{\beta}{\gamma} - \frac{\nu}{\mu})\mathbf{a} = (\frac{\nu}{\lambda} - \frac{\alpha}{\gamma})\mathbf{b}$$
From which it follows that $$ \beta - \frac{\beta}{\gamma} - \frac{\nu}{\mu} = 0 \Rightarrow \gamma = \frac{\nu\gamma}{\beta\mu} + 1 $$ 
and
$$ \frac{\nu}{\lambda} - \frac{\alpha}{\gamma} = 0 \Rightarrow \nu\gamma = \alpha\lambda $$ 
So, once we substitute our values for $ \alpha $ and $ \beta $ $$ \gamma = \frac{\alpha\lambda}{\beta\mu} + 1 = \frac{\mu(\lambda - 1)}{\lambda(\mu - 1)} + 1 $$

3. Tie it all together
By substituting in our values for $\alpha,\beta $ and $ \gamma $ we achieve
$$ 
\begin{align}
\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \times \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \times \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} & = \frac{\alpha - 1}{(\beta - 1)(\gamma - 1)} \\
& = 1
\end{align}
$$

The above proof uses the most fundamental approach of equating two vectors which are the same and solving a pair of simultaneous equations.
