Let $ABCD$ be a rectangle with sides $AB,BC,CD$ and $DA$. Let $K,L$ be the midpoints of the sides $BC,DA$ respectivily. The perpendicular from $B$ to $AK$ hits $CL$ at $M$. Find
$$\frac{[ABKM]}{[ABCL]}$$
1. Let the rectangle \(ABCD\) have sides \(AB = a\) and \(BC = b\). The area of the rectangle is \(A = ab\).

2. The midpoints \(K\) and \(L\) of sides \(BC\) and \(DA\) respectively divide these sides into two equal parts. Therefore, \(BK = \frac{b}{2}\) and \(DL = \frac{a}{2}\).

3. The area of triangle \(ABK\) can be calculated as follows:
   \[
   [ABK] = \frac{1}{2} \times AB \times BK = \frac{1}{2} \times a \times \frac{b}{2} = \frac{ab}{4} = \frac{A}{4}
   \]

4. The line \(AK\) is a diagonal of the rectangle \(ABCD\). Since \(K\) is the midpoint of \(BC\), \(AK\) bisects the rectangle into two equal areas. Therefore, the area of quadrilateral \(AKCL\) is half of the area of the rectangle:
   \[
   [AKCL] = \frac{A}{2}
   \]

5. The perpendicular from \(B\) to \(AK\) intersects \(CL\) at point \(M\). Since \(M\) lies on \(CL\), the area of triangle \(AKM\) is half of the area of quadrilateral \(AKCL\):
   \[
   [AKM] = \frac{1}{2} [AKCL] = \frac{1}{2} \times \frac{A}{2} = \frac{A}{4}
   \]

6. The area of quadrilateral \(ABKM\) is the sum of the areas of triangles \(ABK\) and \(AKM\):
   \[
   [ABKM] = [ABK] + [AKM] = \frac{A}{4} + \frac{A}{4} = \frac{A}{2}
   \]

7. The area of quadrilateral \(ABCL\) is the sum of the areas of triangles \(ABL\) and \(BCL\). Since \(L\) is the midpoint of \(DA\), the area of \(ABL\) is:
   \[
   [ABL] = \frac{1}{2} \times AB \times DL = \frac{1}{2} \times a \times \frac{a}{2} = \frac{a^2}{4}
   \]
   And the area of \(BCL\) is:
   \[
   [BCL] = \frac{1}{2} \times BC \times CL = \frac{1}{2} \times b \times \frac{a}{2} = \frac{ab}{4}
   \]
   Therefore, the area of \(ABCL\) is:
   \[
   [ABCL] = [ABL] + [BCL] = \frac{a^2}{4} + \frac{ab}{4} = \frac{a^2 + ab}{4} = \frac{3A}{4}
   \]

8. The required ratio is:
   \[
   \frac{[ABKM]}{[ABCL]} = \frac{\frac{A}{2}}{\frac{3A}{4}} = \frac{2}{3}
   \]

The final answer is \(\boxed{\frac{2}{3}}\).