Let $BAC$ be a triangle. What can be said about the limiting slope of BC when the $\angle A$ goes to $\pi$.

Question

Let $BAC$ be a triangle. Let a line $L$ of slope $m$ pass through the point $A (z,f(z))$ which is tangent of $f$ at the $x$ coordinate of $A$. Now assume $B$ and $C$ are moving such that $B$ corresponds to $(s,f(s))$ where $s \leq z$ and $C$ corresponds to $(t,f(t))$ where $t \geq z $ and $s, t$ move towards the $x$ coordinate of $A$. Thus, the following condition is satisfied:

1.) The slope of $BA$ and $CA$ tends to slope of $L$

2.) The line $L$ doesn't intersect any sides of triangle $BAC$

Then can we say "using geometry" that the slope of $BC$ tends to slope of $L$

Context

I want to prove the following using the above theorem.

Let $f$ be a function which is differentiable at $z$. This implies that $ \forall \epsilon > 0 \ \exists \delta > 0 \ni \forall s,t \ \ \ \ni z-\delta < s \leq z \leq t < z+\delta \ \ \Bigg|\frac{f(t)-f(s)}{t-s}-f'(z)\Bigg| < \epsilon $

To me the geometry argument seems enough. But my friend said we have to show rigorously that the slope of the segment $BC$ goes to the slope of the line $L$.