Given some function $f(x)=mx+b$ on the interval $[j,k]$, does this prove that $f({{j+k}\over2})$, (the midpoint), is equal to the average value of $f$ on $[j,k]$?
If, using the midpoint formula: $M(a,b)={{a+b}\over2}$ $$A_1={{f(k)+f(j)}\over2}=f({{k+j}\over2})\\={m\over2}(k+j)+b$$ And, using the average formula: $A(a,b)={{\int_a^b{f(x)}}\over{b-a}}$ $$A_2={{\int_j^k{mx+b}}\over{k-j}}\\={{{m\over2}k^2+bk-\big({{m\over2}j^2+bj}\big)}\over{k-j}}\\={{m\over2}(k+j)+b}$$ Then: $A_1=A_2$, which implies that the midpoint is the average.
Yes - this proved the desired result. Your proof could be a little more formal but that does not alter the result.