Let $ f $ be a continuous function in $ [a, b] $ and $ \displaystyle \int_{a} ^ {b} f (t) dt \neq0 $. Prove that at any point $ k \in (0,1) $ there exists a number $ c \in (a, b) $ such that $ \displaystyle \int_{a} ^ {c} f (t) dt = k \int_{a}^{b} f(t) dt $.
I had the intention of solving it by means of the mean value theorem, for example considering that, $$F(x)=\int_{a}^{x}f(t)dt\int_{a}^{b}f(t)dt$$ but I have a problem, because I get that, $$\int_{a}^{c}f(t)dt=\frac{1}{(b-a)f(c)}\left[\int_{a}^{b}f(t)dt\right]^{2}$$
Let $F(x)=\int_a^xf(t)\,\mathrm dt$. Then $F(a)=0$ and $F(b)=\int_a^bf(t)\,\mathrm dt$. Since $k\int_a^bf(t)\,\mathrm dt$ lies between $F(a)$ and $F(b)$, there must be a $c\in(a,b)$ such that$$F(c)\left(=\int_a^cf(t)\,\mathrm dt\right)=k\int_a^bf(t)\,\mathrm dt,$$by the Intermediate Value Theorem.