Step 1: Recall the defining property of an orthogonal matrix.
A matrix \(A\) is orthogonal when its columns are orthonormal, which is the same as saying
\[ A^TA=AA^T=I \]
Every property in this question follows from this single equation.
Step 2: Test statement (A) using the determinant.
Taking the determinant of \(A^TA=I\) gives
\[ \det(A^T)\det(A)=\det(I)=1 \]
Since \(\det(A^T)=\det(A)\), this becomes
\[ (\det A)^2=1 \]
so \(\det(A)=+1\) or \(\det(A)=-1\). Statement (A) is correct.
Step 3: Test statement (B) using eigenvalues.
Suppose \(Av=\lambda v\) for some nonzero eigenvector \(v\), possibly complex. Then
\[ v^*A^TAv=v^*v \]
since \(A^TA=I\). Substituting \(Av=\lambda v\), the left side becomes \(|\lambda|^2v^*v\). So
\[ |\lambda|^2v^*v=v^*v \]
Since \(v\) is nonzero, \(v^*v\neq0\), and dividing gives
\[ |\lambda|^2=1\quad\Rightarrow\quad|\lambda|=1 \]
Statement (B) is correct.
Step 4: Test statement (D) using the norm and inner product.
For any \(x\in R^n\),
\[ \|Ax\|^2=(Ax)^T(Ax)=x^TA^TAx=x^Tx=\|x\|^2 \]
so \(\|Ax\|=\|x\|\) for every \(x\). More generally, for any \(x,y\in R^n\),
\[ (Ax)^T(Ay)=x^TA^TAy=x^Ty \]
This holds for all \(x,y\), not just distinct ones. Statement (D) says exactly this, so it is correct.
Step 5: Test statement (C).
Statement (C) claims \((Ax)^T(Ay)\neq x^Ty\) for all distinct \(x,y\). But Step 4 shows \((Ax)^T(Ay)=x^Ty\) always, for every pair \(x,y\), distinct or not. So statement (C) contradicts a property that always holds, and it is incorrect.
Step 6: Final conclusion.
The correct statements are (A), (B), and (D).
\[
\boxed{\text{(A), (B), and (D) are correct}}
\]