Conditional expectation for a sum of iid random variables: $E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$ I don't really know how to start proving this question.
Let $\xi$ and $\eta$ be independent, identically distributed random variables with $E(|\xi|)$ finite.
Show that
$E(\xi\mid\xi+\eta)=E(\eta\mid\xi+\eta)=\frac{\xi+\eta}{2}$
Does anyone here have any idea for starting this question?

There's a subtle point here, which bothered me the first time I saw this problem.
Henry's answer has the essential idea, which is to use symmetry.  User Did's comment points out that the symmetry comes from the fact that $(\xi, \eta)$ and $(\eta, \xi)$ are identically distributed.  But, straight from the definition of conditional expectation, it isn't clear that symmetry in the joint distributions is enough to get the result.  I ended up having to prove the following lemma:

Lemma. Let $X,Y$ be random variables.  There is a measurable function $f$ such that $E[X\mid Y] = f(Y)$ a.s.  Moreover, if $(X', Y')$ is identically distributed to $(X,Y)$, then $E[X' \mid Y'] = f(Y')$ a.s. for the same function $f$.

Proof.  The existence of $f$ is a consequence of the Doob-Dynkin Lemma.  For the second part, we use the definition of conditional expectation.  $f(Y')$ is clearly $\sigma(Y')$-measurable, so it remains to show that for any $A \in \sigma(Y')$, we have $E[1_A f(Y')] = E[1_A X']$.  Since $A \in \sigma(Y')$, $A = (Y')^{-1}(B)$ for some Borel set $B$ (this fact is part of the proof of Doob-Dynkin).  But since $(X',Y')$ has the same distribution as $(X,Y)$, we get
$$\begin{align*} E[1_A f(Y')] &= E[1_B(Y') f(Y')] \\
&= E[1_B(Y) f(Y)] \\
&= E[1_B(Y) E[X \mid Y]] \\
&= E[1_B(Y) X] && \text{since $1_B(Y)$ is $\sigma(Y)$-measurable}\\
&= E[1_B(Y') X'] \\
&= E[1_A X']
\end{align*}$$ 
as desired.
It is worth noting that the function $f$ is generally not unique.  In particular, we could modify $f$ almost arbitrarily on any set $C \subset \mathbb{R}$ such that $P(Y \in C)=0$.
Also, to address the point in kkk's comment: Just knowing that $\xi, \eta$ are identically distributed is not sufficient.  Here is a counterexample.  Let $\Omega = \{a,b,c\}$ have three outcomes, each with probability $1/3$ (and $\mathcal{F} = 2^\Omega$).  Let $X(a) = 0$, $X(b)=1$, $X(c)=2$; and $Y(a)=1$, $Y(b)=2$, $Y(c)=0$. Thus  $X$ is uniformly distributed on $\{0,1,2\}$, and $Y = X + 1 \bmod 2$, so $Y$ is also uniformly distributed on $\{0,1,2\}$.
Now we have $(X+Y)(a) = 1$, $(X+Y)(b)=3$, $(X+Y)(c)=2$.  So $X+Y$ is a 1-1 function on $\Omega$ and thus $\sigma(X+Y) = \mathcal{F}$, so both $X,Y$ are $\sigma(X+Y)$-measurable.  Thus $E[X\mid X+Y]=X$, $E[Y\mid X+Y]=Y$.  However, $X$, $Y$, and $\frac{X+Y}{2}$ are all different.
