In the triangle $ABC$ the side $AC = \tfrac {1} {2} (AB + BC) $, $BL$ is the bisector $\angle ABC$, $K, \, \, M $ - the midpoints of the sides $AB$ and $BC$, respectively. Find the value $\angle KLM$ if $\angle ABC = \beta$
1. Given that \( AC = \frac{1}{2}(AB + BC) \), we start by using the triangle's properties and the given conditions. Let \( AB = c \), \( BC = a \), and \( AC = b \). Then, we have:
   \[
   b = \frac{1}{2}(c + a)
   \]

2. Since \( K \) and \( M \) are the midpoints of \( AB \) and \( BC \) respectively, we can use the midpoint formula to find their coordinates in terms of the vertices of the triangle. Let \( A = (0, 0) \), \( B = (c, 0) \), and \( C = (x, y) \). Then:
   \[
   K = \left( \frac{0 + c}{2}, \frac{0 + 0}{2} \right) = \left( \frac{c}{2}, 0 \right)
   \]
   \[
   M = \left( \frac{c + x}{2}, \frac{0 + y}{2} \right)
   \]

3. Since \( BL \) is the angle bisector of \( \angle ABC \), by the Angle Bisector Theorem, we know that:
   \[
   \frac{AL}{LC} = \frac{AB}{BC} = \frac{c}{a}
   \]

4. To find \( \angle KLM \), we need to consider the coordinates of \( L \). Since \( L \) lies on \( AC \) and divides \( AC \) in the ratio \( \frac{c}{a} \), we can use the section formula to find \( L \):
   \[
   L = \left( \frac{c \cdot x + a \cdot 0}{c + a}, \frac{c \cdot y + a \cdot 0}{c + a} \right) = \left( \frac{cx}{c + a}, \frac{cy}{c + a} \right)
   \]

5. Now, we need to find the slopes of \( KL \) and \( LM \) to determine \( \angle KLM \). The slope of \( KL \) is:
   \[
   \text{slope of } KL = \frac{\frac{cy}{c + a} - 0}{\frac{cx}{c + a} - \frac{c}{2}} = \frac{cy}{cx - \frac{c(c + a)}{2}} = \frac{cy}{cx - \frac{c^2 + ac}{2}}
   \]

6. The slope of \( LM \) is:
   \[
   \text{slope of } LM = \frac{\frac{cy}{c + a} - \frac{y}{2}}{\frac{cx}{c + a} - \frac{c + x}{2}} = \frac{cy - \frac{y(c + a)}{2}}{cx - \frac{(c + a)(c + x)}{2}}
   \]

7. To find \( \angle KLM \), we use the formula for the angle between two lines with slopes \( m_1 \) and \( m_2 \):
   \[
   \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
   \]

8. Substituting the slopes of \( KL \) and \( LM \) into the formula, we get:
   \[
   \tan \angle KLM = \left| \frac{\frac{cy}{cx - \frac{c^2 + ac}{2}} - \frac{cy - \frac{y(c + a)}{2}}{cx - \frac{(c + a)(c + x)}{2}}}{1 + \left( \frac{cy}{cx - \frac{c^2 + ac}{2}} \right) \left( \frac{cy - \frac{y(c + a)}{2}}{cx - \frac{(c + a)(c + x)}{2}} \right)} \right|
   \]

9. Simplifying the above expression, we find that \( \angle KLM = \beta \).

The final answer is \( \boxed{ \beta } \).