Show that there are $c,d \in (a,b), c\lt d$ such that $\frac{1}{b-a}\int_a^b f=\frac{1}{d-c}\int_c^d f$. Let $f: [a,b] \to R$ be continuous. 
Show that there are $c,d \in (a,b), c\lt d$ such that $\frac{1}{b-a}\int_a^b f=\frac{1}{d-c}\int_c^d f$.
This is my solution. I tried to work from backwards. Letting $F(x)=\int_a^x f$, we want to get the equality $\frac{F(b)-F(a)}{b-a}=\frac{F(d)-F(c)}{d-c}$ for some $c,d\in (a,b)$. 
This is equal to $-F(d)+F(c)+\frac{F(b)-F(a)}{b-a}(d-c)=0$. So let's define $G(x)=F(b)-F(x)-\frac{F(b)-F(a)}{b-a}(b-x)$. 
Then we want $G(d)-G(c)=0$. So since $G(a)=G(b)=0$ and $G$ is continuous on $[a,b]$. Without loss of generality let's assume that the maximum of $G$ is greater than $0$. Say $G(x_1)$ is the maximum. Then for any $r$ such that  $0=G(a)=G(b)\lt r \lt G(x_1)$, by the Intermediate value theorem we can find $c,d$ in $(a,x_1)$ and $(x_1,b)$ respectively, such that $G(c)=G(d)=r$. Hence we have found the desired points.
This solves the problem, however, I feel like I have forced the answer and the solution does not really shed light on the geometric or intuitive meaning of the equality. Is there another way to solve it? Also, what is the meaning behind this equality? I would greatly appreciate any help.
 A: To get some intuition, consider the case $F(b) = 0$ (here using your notation). Assume the maximum value of $F$ occurs at $x_0 \in (a,b),$ with $F(x_0) > 0.$ (Good to draw a picture.) Then by the IVT every value between $0$ and $F(x_0)$ will be taken by $F$ in both of the intervals $(a,x_0)$ and $(x_0,b).$ So for each $y\in (0,F(x_0))$ there exist $c(y) \in (a,x_0), d(y) \in (x_0,b)$ such that $F(c(y)) = y = F(d(y)).$ For each such $y$ we then have
$$\frac{F(d(y))-F(c(y))}{d(y))-c(y)} = 0 = \frac{F(b)-F(a)}{b-a}.$$
Which says there are loads of solutions.
The case where $F$ takes on a minimum value less that $0$ is similar, and of course if $F\equiv 0$ there is nothing to do. The general result follows from the above by looking at $F(x) - l(x),$ where $l$ is the line through $(a,0),(b,F(b)).$
Note that we used only the continuity of $F$ on $[a,b].$ So, back to the original problem, the conclusion holds if we only assume $f$ is Riemann integrable (or even just Lebesgue integrable) on $[a,b].$
