The following is inspired by [this](https://math.stackexchange.com/questions/11487/y-xn-what-is-exyy) recent question on math.stackexchange.
Two standard exercises in conditional expectation are to find ${\rm E}(X\_1|X\_1+X\_2)$ where:
1) $X\_i$, $i=1,2$, are independent ${\rm N}(0,\sigma\_i^2)$ rv's;
2) $X\_i$, $i=1,2$, are independent Poisson($\lambda\_i$) rv's.
The solutions are given by $\frac{{\sigma \_1^2 }}{{\sigma \_1^2 + \sigma \_2^2 }}(X\_1 + X\_2)$ and $\frac{{\lambda \_1 }}{{\lambda \_1 + \lambda \_2 }}(X\_1 + X\_2)$, respectively. A proof for case 1) is given on math.stackexchange. For case 2) we have
${\rm E}(X\_1|X\_1 + X\_2 = n) = \sum\limits\_{k = 0}^n {k{\rm P}(X\_1 = k|X\_1 + X\_2 = n)}.$
A straightforward calculation shows that the right-hand side sum is equal to
$\sum\limits\_{k = 0}^n {k{n \choose k}\bigg(\frac{{\lambda \_1 }}{{\lambda \_1 + \lambda \_2 }}\bigg)^k \bigg(\frac{{\lambda \_2 }}{{\lambda \_1 + \lambda \_2 }}\bigg)^{n - k} },$
which is the expectation of the binomial distribution with parameters $n$ and $\lambda\_1 / (\lambda\_1 + \lambda\_2)$, hence given by $ n \lambda\_1 / (\lambda\_1 + \lambda\_2)$. The result for case 2) is thus proved. In this context, what is common to the normal and Poisson distributions is that both are infinitely divisible (ID). More specifically, 
the characteristic function of $X\_i \sim {\rm N}(0,\sigma\_i^2)$ is given by ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{\sigma \_i^2 ( - z^2 /2)}$, and that of $X\_i \sim {\rm Poisson}(\lambda\_i)$ by ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{\lambda \_i ({\rm e}^{{\rm i}z} - 1)}$. Now, consider integrable, ID, independent rv's $X\_i$, $i=1,2$, with characteristic functions of the form ${\rm E}[{\rm e}^{{\rm i}zX\_i} ] = {\rm e}^{c\_i \psi(z)}$, $c\_i > 0$ (loosely speaking, the characteristic function of an arbitrary ID rv is of that form). In view of the normal and Poisson examples considered above (the former requires somewhat tedious algebra for the solution), and the fact that many important rv's fall into the general category of integrable ID rv's (e.g., gamma rv's), it would be very useful to have the following result: ${\rm E}(X\_1 | X\_1 + X\_2) = \frac{{c \_1 }}{{c\_1 + c\_2 }}(X\_1 + X\_2)$. In fact, I have proved it recently. 
Now to my questions: 1) Have you encountered this result before? 2) Can you provide a rigorous but simple proof of it?
3) Can you provide some intuition? 


EDIT: 1) Here's another interesting example: if $X\_i \sim {\rm Gamma}(c\_i,\lambda)$, $i=1,2$, so that $X\_i$ has density $f\_{X\_i } (x) = \lambda ^{c\_i } {\rm e}^{ - \lambda x} x^{c\_i - 1} /\Gamma (c\_i )$, $x > 0$, then ${\rm E}(X\_1 | X\_1 + X\_2) = \frac{{c\_1 }}{{c\_1 + c\_2 }}(X\_1 + X\_2 )$. 2) It is very instructive to reformulate the result in terms of L\'evy processes: if $X = \{ X(t): t \geq 0 \}$ is an integrable L\'evy process, then ${\rm E}[X(s)|X(t)] = \frac{s}{t}X(t)$, $0 < s < t$.


EDIT: The "direct" solution for the gamma case considered above is now given 
[here](https://math.stackexchange.com/questions/11673/ex-1-x-1-x-2-for-independent-gamma-random-variables). This shows, once more, the effectiveness of the general formula.


EDIT: A complete solution is given in my first (according to date) answer below.


EDIT: An important extension is considered in my second answer below.



$E(X_1 | X_1 + X_2)$, where $X_i$ are (integrable) independent infinitely divisible rv's "of the same type"
Regarding a "rigorous but simple proof" of the relation the OP is interested in, such a proof is, almost completely, already written in the original post. 


To see this, consider independent integrable random variables $X$ and $Y$ and assume that their characteristic functions, defined for every real number $t$, are such that $E({\mathrm e}^{\mathrm{i}tX})=\mathrm{e}^{a\psi(t)}$ and $E(\mathrm{e}^{\mathrm{i}tY})=\mathrm{e}^{b\psi(t)}$ for a given function $\psi$ and given real numbers $a$ and $b$. Let $S=X+Y$. Now, to prove that $$
(a+b)E(X\vert S)=aS,
$$ 
it suffices to show that, for every real number $t$,
$$
(a+b)E(X\mathrm{e}^{\mathrm{i}tS})=aE(S\mathrm{e}^{\mathrm{i}tS}).
$$
Since both sides of the equality can be explicitly written in terms of $a$, $b$, the function $\psi$ and its derivative $\psi'$, the proof is, in a way and modulo some easy computations, already over.


For example, 
$$
E(S\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})+E(Y\mathrm{e}^{\mathrm{i}tY})E({\mathrm e}^{\mathrm{i}tX}),
$$ 
because $X$ and $Y$ are independent. Here, both $E({\mathrm e}^{\mathrm{i}tX})$ and $E({\mathrm e}^{\mathrm{i}tY})$ are already known, and both $E(X{\mathrm e}^{\mathrm{i}tX})$ and $E(Y{\mathrm e}^{\mathrm{i}tY})$ are derivatives of the former with respect to $(\mathrm{i}t)$. Hence, 
$$
E(S\mathrm{e}^{\mathrm{i}tS})=-\mathrm{i}(a+b)\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$
Likewise, 
$$
E(X\mathrm{e}^{\mathrm{i}tS})=E(X\mathrm{e}^{\mathrm{i}tX})E({\mathrm e}^{\mathrm{i}tY})=-\mathrm{i}a\psi'(t)\mathrm{e}^{(a+b)\psi(t)}.
$$ 
Comparing these two formulas, we are done.


(If this helps, one can note that the signs of $a$ and $b$ must be the same, in the sense that $ab>0$ or that $X$ or $Y$ must be $0$ with full probability.)


