Let $ k$ be a positive real number. Triangle ABC is acute and not isosceles, O is its circumcenter and AD,BE,CF are the internal bisectors. On the rays AD,BE,CF, respectively, let points L,M,N such that $ \frac {AL}{AD} \equal{} \frac {BM}{BE} \equal{} \frac {CN}{CF} \equal{} k$. Denote $ (O_1),(O_2),(O_3)$ be respectively the circle through L and touches OA at A, the circle through M and touches OB at B, the circle through N and touches OC at C. 
1) Prove that when $ k \equal{} \frac{1}{2}$, three circles $ (O_1),(O_2),(O_3)$ have exactly two common points, the centroid G of triangle ABC lies on that common chord of these circles. 
2) Find all values of k such that three circles $ (O_1),(O_2),(O_3)$ have exactly two common points
1. **Prove that when \( k = \frac{1}{2} \), three circles \((O_1), (O_2), (O_3)\) have exactly two common points, the centroid \( G \) of triangle \( ABC \) lies on that common chord of these circles.**

   Let's denote the centroid of triangle \( ABC \) as \( G \). The centroid \( G \) divides each median in the ratio \( 2:1 \). Since \( k = \frac{1}{2} \), the points \( L, M, N \) are the midpoints of \( AD, BE, CF \) respectively.

   - **Step 1:** Consider the circle \((O_1)\) passing through \( L \) and tangent to \( OA \) at \( A \). Since \( L \) is the midpoint of \( AD \), \( AL = \frac{1}{2}AD \). The circle \((O_1)\) is the Apollonian circle of \( A \) and \( D \) with ratio \( \frac{1}{2} \).

   - **Step 2:** Similarly, the circle \((O_2)\) passing through \( M \) and tangent to \( OB \) at \( B \) is the Apollonian circle of \( B \) and \( E \) with ratio \( \frac{1}{2} \).

   - **Step 3:** The circle \((O_3)\) passing through \( N \) and tangent to \( OC \) at \( C \) is the Apollonian circle of \( C \) and \( F \) with ratio \( \frac{1}{2} \).

   - **Step 4:** Since \( L, M, N \) are the midpoints of \( AD, BE, CF \) respectively, and the centroid \( G \) divides each median in the ratio \( 2:1 \), \( G \) lies on the common chord of these circles. This is because the centroid \( G \) is the intersection of the medians, and the medians are the radical axes of the circles \((O_1), (O_2), (O_3)\).

   Therefore, when \( k = \frac{1}{2} \), the three circles \((O_1), (O_2), (O_3)\) have exactly two common points, and the centroid \( G \) of triangle \( ABC \) lies on that common chord.

2. **Find all values of \( k \) such that three circles \((O_1), (O_2), (O_3)\) have exactly two common points.**

   - **Step 1:** We need a lemma: Given \( \triangle ABC \) and an arbitrary point \( S \). Let \( D \equiv AP \cap BC \), \( E \equiv BP \cap CA \), \( F \equiv CP \cap AB \), \( G \equiv EF \cap BC \), \( H \equiv FD \cap CA \), \( I \equiv ED \cap AB \). Let \( A_k \in BC \), \( B_k \in CA \), \( C_k \in AB \) which satisfy: \(\dfrac{\overline{DA_k}}{\overline{DG}} = \dfrac{\overline{EB_k}}{\overline{EH}} = \dfrac{\overline{FC_k}}{\overline{FI}} = k\). Prove that: \( A_k, B_k, C_k \) are collinear \(\Leftrightarrow k = 1 \) or \( k = \dfrac{1}{2} \).

   - **Step 2:** We have: \(\dfrac{\overline{GB}}{\overline{GC}} \cdot \dfrac{\overline{HC}}{\overline{HA}} \cdot \dfrac{\overline{IA}}{\overline{IB}} = - \dfrac{\overline{DB}}{\overline{DC}} \cdot \dfrac{\overline{EC}}{\overline{EA}} \cdot \dfrac{\overline{FA}}{\overline{FB}} = 1\). Then: \( G, H, I \) are collinear.

   - **Step 3:** Let \( M, N, P \) be the midpoints of \( AG, BH, CI \) respectively. So: \( M, N, P \) lie on the Newton line of the complete quadrilateral \( ABMN.CP \). Hence: \( k = 1 \) or \( k = \dfrac{1}{2} \) satisfy the problem.

   - **Step 4:** Suppose there exists \( k \ne 1 \), \( k \ne \dfrac{1}{2} \) that satisfies the problem. It's easy to prove that: \(\dfrac{\overline{A_kM}}{\overline{A_kG}} = \dfrac{\overline{B_kN}}{\overline{B_kH}} = \dfrac{\overline{C_kP}}{\overline{C_kI}} = \dfrac{1 - 2k}{2 - 2k}\). Then: \(\overline{M, N, P} \parallel \overline{A_k, B_k, C_k} \parallel \overline{G, H, I}\) or \(\dfrac{\overline{HG}}{\overline{HI}} = \dfrac{\overline{NM}}{\overline{NP}}\).

   - **Step 5:** But: \(\dfrac{\overline{DM}}{\overline{DG}} = \dfrac{\overline{EN}}{\overline{EH}} = \dfrac{\overline{FP}}{\overline{FI}} = \dfrac{1}{2}\) and \( D, E, F \) aren't collinear, so: \(\dfrac{\overline{HG}}{\overline{HI}} = \dfrac{\overline{NM}}{\overline{NP}}\) cannot happen. Combine with: \(\overrightarrow{NM} = \dfrac{1}{2} (\overrightarrow{HG} + \overrightarrow{ED})\), \(\overrightarrow{NP} = \dfrac{1}{2} (\overrightarrow{HI} + \overrightarrow{EF})\) and \( D, E, F \) aren't collinear, then: \(\overline{M, N, P} \parallel \overline{A_k, B_k, C_k} \parallel \overline{G, H, I}\) cannot happen.

   - **Step 6:** Hence: There don't exist \( k \ne 1 \), \( k \ne \dfrac{1}{2} \) that satisfy the problem. Therefore: \( A_k, B_k, C_k \) are collinear \(\Leftrightarrow k = 1 \) or \( k = \dfrac{1}{2}\).

The final answer is \( \boxed{ k = 1 } \) or \( k = \frac{1}{2} \).