Intuition for $RSS_2 - RSS_1$ having chi-square distribution in F-test for linear models In https://en.wikipedia.org/wiki/F-test#Regression_problems, an application of the F-statistic to comparing linear models is given:

Consider two models, 1 and 2, where model 1 is 'nested' within model 2. Model 1 is the restricted model, and model 2 is the unrestricted one. That is, model 1 has p1 parameters, and model 2 has p2 parameters, where p1 < p2, and for any choice of parameters in model 1, the same regression curve can be achieved by some choice of the parameters of model 2.
[...]
If there are n data points to estimate parameters of both models from, then one can calculate the F statistic, given by

$$\frac{RSS_1-RSS_2}{p_2-p_1} / \frac{RSS_2}{n - p_2}$$
I understand why $RSS_2$ is chi-square distributed with degrees of freedom $n - p_2$: this is because in linear regression we project the data onto a $p_2$ dimensional space, leaving the residual error to reside in an $n-p_2$ dimensional space.
My question is why is $RSS_1 - RSS_2$ Chi-square distributed with $p_2-p_1$ degrees of freedom? It is true that $RSS_1$ and $RSS_2$ are chi-square distributed, but in general, I think the difference of two chi-squared distributions is not chi-square distributed. So this is a confusing claim to me.
I tried looking at the proof here: Proof that F-statistic follows F-distribution but it went above my head. I'm not looking for a completely rigorous proof, but I am looking for an explanation that I can grok that doesn't also wrongly (?) imply that the difference of any two chi-squared distributions is chi-squared, probably somehow using the fact that the models are nested.

Below we assume we have a model $y=Xb+e$ where the components of vector $e$ are distributed iid $N(0, \sigma^2)$ and that $H$ is the projection matrix onto the range of $X$. Use a subscript of $r$ to denote corresponding quantities of the restricted model and use $\hat{e} = (I-H)y$ to mean the estimated value of the unknown $e$.
We will make use of some facts about projections shown in the Appendix at the end.
Note that
$RSS = \hat{e}'\hat{e} = y'(I-H)y = e'(I-H)e$
where we have used the fact that
$(I-H)y = (I-H)(Xb+e) = (I-H)e$
Now applying that to both $RSS$ and $RSS_r$ we have that
$RSS_r - RSS = e'(I-H_r)e - e'(I-H)e = e'(H-H_r)e$
$H$ and $H_r$ are projections and their difference is a projection too because the space that $H_r$ projects onto is nested within the space associated with $H$ by our assumptions.  Thus $RSS_r - RSS$ is of the form $e'Pe$ where $P = H-H_r$ is an orthogonal projection.
Now it is known that if the components of vector $x$ are iid $N(0, 1)$, which is the case for $e/\sigma$ by assumption, and $Q$ is any orthogonal projection that $x'Qx$ is chi-squared with degrees of freedom equal to the dimension of the space onto which $Q$ projects (and is also equal to the rank of $Q$ and also equal to $trace(Q)$).  Thus $(RSS-RSS_r)/\sigma^2$ is chi-squared.
Appendix
An orthogonal projection matrix $Q$ is a matrix which satisfies $Q = Q' = Q^2$.  That is it is symmetric and idempotent.  This implies that $||Qx||^2 = x'Q'Qx = x'Qx$ where $||.||^2$ means squared length.
If $Q$ is an orthogonal projection then so is $I-Q$.
The range of a matrix is the set of values it maps to. An orthogonal projection is said to project onto its range.
If an orthogonal projection $Q$ projects onto the range of matrix $M$ then $QM=M$ and $(I-Q)M = 0$.
If $Q$ and $Q_0$ are orthogonal projections such that $Q_0$ projects onto a subspace of the set that $Q$ projects onto then $Q$ and $Q_0$ commute and $Q-Q_0$ is an orthogonal projection too.  Also $rank(Q-Q_0) = rank(Q) - rank(Q_0)$.
