Answer
The slope of a cubic Bézier curve at $P_3$ is equal to the slope of the segment $\overline{P_2 P_3}$, therefore the cubic Bézier curve is tangent to the segment $\overline{P_2 P_3}$ at $P_3$.
Work Step by Step
A cubic Bézier curve is given by $c(t)=(x(t),y(t))$, where
$x(t)=a_0 (1-t)^3+3 a_1 t (1-t)^2+3 a_2 t^2 (1-t)+a_3 t^3$,
$y(t)=b_0 (1-t)^3+3 b_1 t (1-t)^2+3 b_2 t^2 (1-t)+b_3 t^3$.
Using Eq. (8) the slope of the tangent line is given by
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{-3 b_0 (1-t)^2+3 b_1 (1-t)^2-6 b_1 t (1-t)+6 b_2 t (1-t)-3 b_2 t^2+3 b_3 t^2}{-3 a_0 (1-t)^2+3 a_1 (1-t)^2-6 a_1 t (1-t)+6 a_2 t (1-t)-3 a_2 t^2+3 a_3 t^2}$
At $P_3$, $t=1$. So, $\frac{dy}{dx}|_{t=1}=\frac{-3 b_2+3 b_3}{-3 a_2+3 a_3}=\frac{b_3-b_2}{a_3-a_2}$.
Since $P_2=(a_2,b_2)$ and $P_3=(a_3,b_3)$, the slope of the segment $\overline{P_2 P_3}$ is $\frac{b_3-b_2}{a_3-a_2}$. Since the two slopes are equal, a cubic Bézier curve is tangent to the segment $\overline{P_2 P_3}$ at $P_3$.
