I already proved it, but it was really laborious. I am wondering if any one has a shorter proof?
Write $A = [a_{ik}]$ and let $\overline{A}_{rs} = [c_{ik}]$ denote the minor with row $r$ and column $s$ removed. Then, we have: $$c_{ik} = \begin{cases} \, a_{ik} & i < r, k < s \\ \, a_{i(k+1)} \, \, \, \text{ when } & i < r, s \le k \\ \, a_{(i+1)k} & r \le i, k < s \\ \, a_{(i+1)(k+1)} & r \le i, s \le k \end{cases}$$ Suppose that $\overline{A}_{rs}$ is not upper triangular $\rightarrow c_{ik} \ne 0$ with $i > k$. Clearly, this is not possible in the first, third, and fourth case. In the case with $i < r, s \le k$, $a_{i(k+1)} \ne 0$. Since $i > k$ and $A$ is upper triangular, $a_{i(k+1)} = a_{ii}$. This proves that if $c_{ik}$ is nonzero, $k = i - 1$. Furthermore: $$s \le k = i - 1 < i < r$$ Now, take a minor $\overline{A}_{rs}$ with $r < s$. Suppose that it is not upper triangular; $c_{ik} \ne 0$ for some $i > k \rightarrow s \le i -1 < i < r$, a contradiction. Moreover, $c_{rr} = a_{(r+1)r} = 0$. Hence, $\overline{A}_{rs}$ is upper triangular with at least one zero in its diagonal entries. The cofactor $(-1)^{r+s} \det \overline{A}_{rs} = 0$.
$\Gamma$, the matrix made up of the cofactors of $A$ is a lower triangular. It follows that $\text{Adj } A = \Gamma^t$ is upper triangular.
Suppose that $(e_1,...,e_n)$ is the orthogonal base. $A$ is upper triangular means $<A(e_i),e_j) =0$, $j>i$. This is equivalent to $<e_i,A^*(e_j)>=0$ $j>i$ so $A^*$ is lower triangular.