The fraction $\frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac}$ is (with suitable restrictions of the values of a, b, and c):
$\text{(A) irreducible}\qquad$
$\text{(B) reducible to negative 1}\qquad$
$\text{(C) reducible to a polynomial of three terms}\qquad$
$\text{(D) reducible to} \frac{a-b+c}{a+b-c}\qquad$
$\text{(E) reducible to} \frac{a+b-c}{a-b+c}$
Use the commutative property to get
\[\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\]
Factor perfect square trinomials to get
\[\frac{(a+b)^2-c^2}{(a+c)^2-b^2}\]
Factor difference of squares to get
\[\frac{(a+b+c)(a+b-c)}{(a+b+c)(a-b+c)}\]
Cancel out like terms (with suitable restrictions of a, b, c) to get
\[\frac{a+b-c}{a-b+c}\]
The answer is $\boxed{\textbf{(E)}}$.