Simple proof of "maximum number of right angles in a convex $n$-gon is 3 for $n\geq 5$" for a 8th grade student?

Question

I know a proof of "maximum number of right angles in a convex $n$-polygon is 3 for $n\geq 5$" as follows:

Suppose $k$ is the number of right angles. Then $180(n-2)-90k$ is the sum of other $n-k$ interior angles. Now we can perform these $n-k$ angles such that all have equal angle i.e. equal to their average $\frac{180(n-2)-90k}{n-k}=\frac{180(n-k)+90k-360}{n-k}=180+\frac{90(k-4)}{n-k}$ that $\frac{90(k-4)}{n-k}$ must be negative since each angle $<180$ so $90(k-4)<0$ and $k<4$.

But this proof is a bit cumbersome for 8$^{th}$ grade school student. (The bolded part is also dubious to me and hard to accept it! Let alone the 8$^{th}$ grade school student). Is there any more simple argument?

Answer 1

Sum of exterior angles of a convex n-gon is $360^{\circ} = 4\cdot 90^{\circ}$. We conclude for $n > 4$, at most $3$ right angles are allowed. In that case, sum of rest $n-3$ exterior angles is $90^{\circ}$.