A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad
\textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad
\textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad
\textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad
\textbf{(E)}\ x=\dfrac{1}{2}b$
1. Let the base of the rectangle be \( q \). Given that the altitude of the rectangle is \( x \), we have \( x = \frac{1}{2}q \). 
2. Since the rectangle is inscribed in the triangle and its base coincides with the base \( b \) of the triangle, we know that \( \overline{DE} \parallel \overline{AB} \). Therefore, \( \triangle CDE \sim \triangle CAB \) by AA similarity criterion.
3. Let \( DG = JH = EF = x \). The height of the triangle is \( h \), so \( JC = CH - JH = h - x \).
4. The base of the rectangle \( GF = q = 2x \).
5. Using the similarity of triangles \( \triangle CDE \) and \( \triangle CAB \), we can set up the ratio of corresponding sides:
   \[
   \frac{h - x}{h} = \frac{2x}{b}
   \]
6. Cross-multiplying to solve for \( x \):
   \[
   (h - x)b = 2xh
   \]
   \[
   bh - bx = 2xh
   \]
7. Rearranging the terms to isolate \( x \):
   \[
   bh = 2xh + bx
   \]
   \[
   bh = x(2h + b)
   \]
   \[
   x = \frac{bh}{2h + b}
   \]

The final answer is \(\boxed{x = \frac{bh}{2h + b}}\)