In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle  CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?
1. **Reflecting Points and Establishing Cyclic Quadrilateral:**
   - Reflect point \( D \) across \( H \) to get point \( E \), and reflect point \( E \) across \( A \) to get point \( F \). 
   - Since \( \angle ABC = 90^\circ \) and \( \angle BAC = \angle CAD \), triangle \( ABC \) is right-angled at \( B \) and \( AC = AD \).
   - By reflecting, we ensure that \( B \) lies on line \( EF \) and quadrilateral \( DCEF \) is cyclic with diameter \( EF \).

2. **Simson Line and Perpendicular Foot:**
   - Let \( G \) be the foot of the perpendicular from \( C \) to \( DF \). By the properties of the Simson line, \( G \) lies on \( BH \).
   - Since \( DH \) is the altitude of triangle \( ACD \), \( DH \perp AC \) and \( DH \perp CD \).

3. **Rectangle Formation and Bisecting Segment:**
   - Since \( DH \) is perpendicular to \( CD \) and \( G \) is the foot of the perpendicular from \( C \) to \( DF \), quadrilateral \( DHCG \) forms a rectangle.
   - In a rectangle, the diagonals bisect each other. Therefore, \( HG \) bisects \( CD \).

Conclusion:
Since \( HG \) bisects \( CD \), the line \( BH \) divides the segment \( CD \) into two equal parts.

The final answer is \(\boxed{1:1}\)