Step 1: Write the characteristic equation of A.
For a $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the eigenvalues solve
\[
\lambda^2 - (a+d)\lambda + (ad-bc) = 0
\]
Step 2: Use the given condition to simplify the determinant.
The condition $a+b=c+d$ gives $c = a+b-d$. Substitute this into $ad-bc$:
\[
ad - bc = ad - b(a+b-d) = ad - ab - b^2 + bd
\]
Group the terms in pairs:
\[
= d(a+b) - b(a+b) = (a+b)(d-b)
\]
So the determinant of $A$ factors neatly as $(a+b)(d-b)$.
Step 3: Factor the characteristic equation using this determinant.
The characteristic equation becomes
\[
\lambda^2 - (a+d)\lambda + (a+b)(d-b) = 0
\]
Check that $\lambda = a+b$ and $\lambda = d-b$ are roots: their sum is $(a+b)+(d-b) = a+d$, matching the coefficient of $\lambda$, and their product is $(a+b)(d-b)$, matching the constant term. So the equation factors as
\[
\big(\lambda-(a+b)\big)\big(\lambda-(d-b)\big) = 0
\]
Step 4: Read off the two exact eigenvalues.
The only two eigenvalues of $A$ are $a+b$ and $d-b$. There is no algebraic way to get $d+b$ out of this factorisation unless $b=0$, which is excluded by the question. So the statement claiming $d+b$ is an eigenvalue does not hold in general.
\[
\boxed{\text{Eigenvalues are } a+b \text{ and } d-b, \text{ not } d+b}
\]