The tangents at $B$ and $A$ to the circumcircle of an acute angled triangle $ABC$ meet the tangent at $C$ at $T$ and $U$ respectively.  $AT$ meets $BC$ at $P$, and $Q$ is the midpoint of $AP$; $BU$ meets $CA$ at $R$, and $S$ is the midpoint of $BR$.  Prove that $\angle ABQ=\angle BAS$.  Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.
1. **Define the points and tangents:**
   - Let \( A_1, B_1, C_1 \) be the midpoints of \( BC, CA, AB \) respectively.
   - Note that \( Q \) is the foot of the \( A \)-symmedian in \( \triangle AB_1C_1 \).
   - Let \( D = BQ \cap A_1B_1 \) and \( E = AS \cap A_1B_1 \).

2. **Determine the ratios:**
   - Since \( Q \) is the foot of the \( A \)-symmedian, we have:
     \[
     \frac{B_1D}{C_1D} = \frac{b^2}{c^2}
     \]
   - Therefore, \( B_1D = \frac{b^2}{b^2 + c^2} \cdot A_1B_1 \).

3. **Similarity of triangles:**
   - By the properties of the symmedian, \( \triangle DB_1A \sim \triangle CAB \).
   - Similarly, \( A_1E = \frac{a^2}{a^2 + c^2} \cdot A_1B_1 \) and \( \triangle EA_1B \sim \triangle CBA \).

4. **Angles in the isosceles trapezoid:**
   - Since \( \angle D = \angle E = \angle C \), quadrilateral \( ABED \) is an isosceles trapezoid.
   - This implies \( \angle DBA = \angle EAB \).

5. **Conclusion for the first part:**
   - Therefore, \( \angle QBA = \angle SAB \), as desired.

6. **Maximizing the angle:**
   - Let \( BD \) and \( AE \) meet at \( X \) and let \( N \) be the projection from \( C_1 \) onto \( A_1B_1 \).
   - We have:
     \[
     \frac{C_1X}{XN} = \frac{AB}{DE} = \frac{2c^2}{a^2 + b^2 + c^2}
     \]
   - To maximize \( \angle XBA \), we need to maximize the ratio \( \frac{C_1X}{XN} \), which amounts to minimizing \( a^2 + b^2 \).

7. **Minimizing \( a^2 + b^2 \):**
   - As \( C \) varies, this happens when \( CA = CB \), implying \( a = b \).

8. **Calculating the tangent:**
   - Let \( h_c \) be the length of the altitude from \( C \):
     \[
     C_1X = \frac{MX}{MX + XN} \cdot \frac{h_c}{2} = \frac{2c^2}{a^2 + b^2 + 3c^2} \cdot \frac{h_c}{2} = \frac{c^2 \sqrt{4a^2 - c^2}}{4a^2 + 6c^2}
     \]
   - Therefore:
     \[
     \tan \angle XBA = \frac{\sqrt{c^2(4a^2 - c^2)}}{2a^2 + 3c^2}
     \]

9. **Maximizing the tangent:**
   - By AM-GM inequality:
     \[
     \frac{\sqrt{7c^2(4a^2 - c^2)}}{\sqrt{7}(2a^2 + 3c^2)} \le \frac{2a^2 + 3c^2}{\sqrt{7}(2a^2 + 3c^2)} = \frac{1}{\sqrt{7}}
     \]
   - Equality holds when \( 7c^2 = 4a^2 - c^2 \), implying \( a : b : c = \sqrt{2} : \sqrt{2} : 1 \).

The final answer is \( \boxed{ a : b : c = \sqrt{2} : \sqrt{2} : 1 } \)