In a circumference with center $ O$ we draw two equal chord $ AB\equal{}CD$ and if $ AB \cap CD \equal{}L$ then $ AL>BL$ and $ DL>CL$
We consider $ M \in AL$ and $ N \in DL$ such that $ \widehat {ALC} \equal{}2 \widehat {MON}$ 
Prove that the chord determined by extending $ MN$ has the same as length as both $ AB$ and $ CD$
1. **Given:**
   - Two equal chords \( AB \) and \( CD \) in a circle with center \( O \).
   - \( AB \cap CD = L \) such that \( AL > BL \) and \( DL > CL \).
   - Points \( M \in AL \) and \( N \in DL \) such that \( \angle ALC = 2 \angle MON \).

2. **Objective:**
   - Prove that the chord determined by extending \( MN \) has the same length as both \( AB \) and \( CD \).

3. **Initial Setup:**
   - Let \( OM \) and \( ON \) intersect the line segment \( BC \) at points \( E \) and \( F \) respectively.
   - Let \( P \) and \( Q \) be the projections of \( O \) onto \( AB \) and \( CD \) respectively.

4. **Properties of the Chords and Projections:**
   - Since \( AB = CD \) and \( ABCD \) is cyclic, \( ABCD \) forms an isosceles trapezoid.
   - \( BL = CL \implies AL = DL \).
   - \( \angle BAD = \angle CDA = \angle DCB = \angle ABC \).

5. **Perpendicular Projections:**
   - \( OP \perp AB \) and \( OQ \perp CD \) imply \( AP = PB \) and \( QD = QC \).
   - \( AL = DL \) implies \( PL = QL \).

6. **Angles and Parallel Lines:**
   - \( \angle LPQ = \angle LQP = \angle ABC \) implies \( BC \parallel PQ \).
   - \( OL \perp PQ \) implies \( \angle MOP = \angle NOL \).
   - \( OL \perp BC \) implies \( \angle DCB = \angle NOE \).

7. **Cyclic Quadrilaterals:**
   - \( CENO \) is cyclic, and similarly, \( BFMO \) is cyclic.
   - Since \( OB = OC \), \( \angle OBC = \angle OCB \) implies \( \angle OCN = \angle OBM = \angle OEN = \angle OFM \).

8. **Cyclic Properties and Angles:**
   - \( MEFN \) is cyclic, implying \( \angle OFE = \angle OMN \).
   - \( \angle MOP = \angle NOL \) implies \( \angle OFE = \angle OMP \).

9. **Chord Lengths:**
   - \( \angle OMP = \angle OMN \) implies \( \angle OMX = \angle OMB \) if \( MN \) cuts the circle at \( X \) and \( Y \).
   - Let the angle bisector of \( \angle BOX \) cut \( XM \) at \( K \), and assume \( K \) lies between \( X \) and \( M \).

10. **Contradiction and Conclusion:**
    - \( \angle OMX + \angle KOM = \angle OKX = \angle OKB \) implies \( \angle OKB > \angle OMX = \angle OMB \), which is a contradiction.
    - Thus, \( M = K \) implies \( MB = MX \), making \( AXBY \) an isosceles trapezoid.
    - Therefore, \( XY = AB \).

The final answer is \( \boxed{ XY = AB } \).