Could someone tell me if my proof is valid? $f$ is real function continuous at $[a,b]$ such that $f''(x)$ exists $\forall x\in(a,b)$

Question

Assume that line which connects points $A=(a,f(a))$ and $B=(b,f(b))$ intersects graph of the function $f$ at point $(c,f(c))$ where $a<c<b$. Prove that exist two different points $c_0,c_1\in(a,b)$ such that $f'(c_0)=f'(c_1)$.

$\triangle:$

According to MVT $\exists c\in(a,b)$ that $f'(c)$ is direction coefficient of $AB$. Let there be $M\in(a,b)$.Let $Y=(M,f(M))$. If $c_0\in(a,M)$ and $c_1\in(M,b)$ then $f'(c_0)$ is direction coefficient of line that passes through $AY$ and $f'(c_1)$ is direction coefficient of line $YB$. But since those two line are same we have that their coefficients are same; that is $f(c_0)=f(c_1)$

Answer 1

Consider $f:(a,c)\to \mathbb{R}$ and $f:(c,b)\to \mathbb{R}$ then there is $c_0\in(a,c)$ and $c_1\in(c,b)$ such that $f^{\prime}(c_0)=\frac{f(c)-f(a)}{c-a}=\frac{f(b)-f(a)}{b-a}=\frac{f(b)-f(c)}{b-a}=f^{\prime}(c_1)$ by MVT.

Answer 2

The only problem is that you said $M$ might be any point between $(a, b)$, so you have no reason at all to state $AY$ and $YB$ have the same slope. The assumption only states that there is some point in $(a, b)$ that has this property.