This question relates to this one: Understanding a Substep of the Proof for the Law of Total Variance
It is written there, that we should be able to define conditional variance as $Var[X|\mathcal{F}]=E[X^2|\mathcal{F}]-(E[X|\mathcal{F}])^2$. But in order to pass to this definition from usual one, i.e. $$ Var[X|\mathcal{F}]=E[(X-E[X|\mathcal{F}])^2|\mathcal{F}] $$ I need to know that $$ E[XE[X|\mathcal{F}]|\mathcal{F}]=(E[X|\mathcal{F}])^2. $$ Seems plausible, but how I can see that?
$E(XY|\mathcal F)=YE(X|\mathcal F)$ whenever $Y$ is measurable w.r.t. $\mathcal F$ and $E|XY| <\infty$. [Here $X$ and $Y$ have finite second moments so $E|XY| \leq \sqrt {EX^{2}} \sqrt {EY^{2}} <\infty$]. Take $Y=E(X|\mathcal F)$ in this formula.