An altitude $h$ of a triangle is increased by a length $m$. How much must be taken from the corresponding base $b$ so that the area of the new triangle is one-half that of the original triangle? 

${{ \textbf{(A)}\ \frac{bm}{h+m}\qquad\textbf{(B)}\ \frac{bh}{2h+2m}\qquad\textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad\textbf{(D)}\ \frac{b(m+h)}{2m+h} }\qquad\textbf{(E)}\ \frac{b(2m+h)}{2(h+m)} } $
1. Let \( x \) be the length that must be taken from \( b \) so that the area of the new triangle is one-half that of the original triangle.
2. The area of the original triangle is given by:
   \[
   \text{Area}_{\text{original}} = \frac{1}{2} b h
   \]
3. The new altitude is \( h + m \), and the new base is \( b - x \). The area of the new triangle is:
   \[
   \text{Area}_{\text{new}} = \frac{1}{2} (b - x)(h + m)
   \]
4. According to the problem, the area of the new triangle should be one-half of the area of the original triangle:
   \[
   \frac{1}{2} (b - x)(h + m) = \frac{1}{2} \left( \frac{1}{2} b h \right)
   \]
5. Simplify the right-hand side:
   \[
   \frac{1}{2} (b - x)(h + m) = \frac{1}{4} b h
   \]
6. Multiply both sides by 2 to clear the fraction:
   \[
   (b - x)(h + m) = \frac{1}{2} b h
   \]
7. Expand the left-hand side:
   \[
   b h + b m - x h - x m = \frac{1}{2} b h
   \]
8. Rearrange the equation to isolate \( x \):
   \[
   b h + b m - \frac{1}{2} b h = x h + x m
   \]
9. Combine like terms:
   \[
   \frac{1}{2} b h + b m = x (h + m)
   \]
10. Solve for \( x \):
    \[
    x = \frac{\frac{1}{2} b h + b m}{h + m}
    \]
11. Factor out \( b \) from the numerator:
    \[
    x = \frac{b \left( \frac{1}{2} h + m \right)}{h + m}
    \]
12. Simplify the expression:
    \[
    x = \frac{b (2m + h)}{2(h + m)}
    \]

Thus, our answer is \(\boxed{\textbf{(E)}}\).