We have three points: $ACB$ in clockwise order Or Vector \[ \vec{CA} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad % \vec{CB} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} % \] We can use the sign of determinant to check the order of three points: \[ M = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \det M = 1 > 0 \\ \] If we check points: $A, C, B'$ in counter clockwise then we have following two vectors: \[ \vec{CA} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad % \vec{CB'} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \] we can form a matrix: \[ M' = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \\ \det M' = -1 < 0 \] We can conclude the following:
If the determinant is less than 0 then the three points are counter clockwise or CCW
If the determinant is greater than 0 then the three points are clockwise or CW

From above picture:
We can come up some intuition guessing:
$\angle D\color{red}{F}B + \angle A\color{red}{F}D + \angle A\color{red}{F}B > 360$
$\angle C\color{red}{D}B + \angle A\color{red}{D}C + \angle B\color{red}{D}A = 360$
$\angle C\color{red}{E}B + \angle A\color{red}{E}C + \angle A\color{red}{E}B = 360$
Note: if the point $E$ is the same as $A$, $B$ or $C$, then there is "divided by zero" problem
1. Check $E$ is diffed from $A, B, C$
2. Check if they are collinear with segments: $\overline{AB}, \overline{BC}, \overline{AB}$
3. Compute the angles with dot product.
\begin{equation} \begin{aligned} \cos \angle CDB &= \frac{\vec{DC} \circ \vec{DB}}{ | \vec{DC} | | \vec{DB}|} \\ \cos \angle ADC &= \frac{\vec{DA} \circ \vec{DC}}{ | \vec{DA} | | \vec{DC}|} \\ \cos \angle BDA &= \frac{\vec{DB} \circ \vec{DA}}{ | \vec{DB} | | \vec{DA}|} \\ 2\pi &= \cos \angle CDB + \cos \angle ADC + \cos \angle BDA \\ \end{aligned} \end{equation}