ABCD is a square L and K are two point on AB s.t. BO = BK , AO= AL , O is the intersection point of diagonal .If `angle LOK= Q ` find tan Q

Answer

To solve the problem, we need to find the value of \( \tan Q \) where \( Q \) is the angle \( LOK \) in the square \( ABCD \) with points \( L \) and \( K \) on side \( AB \) such that \( BO = BK \) and \( AO = AL \).

### Step-by-Step Solution:

1. **Draw the Square**:
- Let \( ABCD \) be a square with vertices \( A(0, 0) \), \( B(1, 0) \), \( C(1, 1) \), and \( D(0, 1) \).
- The intersection point of the diagonals \( O \) will be at \( (0.5, 0.5) \).

**Hint**: Remember that the diagonals of a square bisect each other at right angles.

2. **Identify Points \( L \) and \( K \)**:
- Since \( AO = AL \) and \( BO = BK \), we can place points \( L \) and \( K \) on \( AB \).
- Let \( AO = AL = x \) and \( BO = BK = x \). Thus, \( L \) will be at \( (x, 0) \) and \( K \) will be at \( (1-x, 0) \).

**Hint**: Use the property of equal segments to position points accurately.

3. **Determine Angles**:
- Since \( O \) is the midpoint of the square, the angle \( AOB \) is \( 90^\circ \).
- In triangle \( BOK \), since \( BO = BK \), it is an isosceles triangle. Let the angles at \( O \) be \( x \).
- Therefore, \( 2x + 90^\circ = 180^\circ \) gives \( 2x = 90^\circ \) or \( x = 45^\circ \).

**Hint**: Use the angle sum property of triangles to find the unknown angles.

4. **Find Angle \( KOL \)**:
- In triangle \( KOL \), we have \( KOL + x + x = 180^\circ \).
- Substituting \( x = 45^\circ \), we get \( KOL + 90^\circ = 180^\circ \), thus \( KOL = 90^\circ \).

**Hint**: Again, apply the angle sum property for triangle \( KOL \).

5. **Calculate \( \tan Q \)**:
- Since \( Q = KOL \) and \( KOL = 45^\circ \), we find \( \tan Q = \tan 45^\circ = 1 \).

**Hint**: Remember the basic trigonometric values for common angles.

### Final Answer:
Thus, the value of \( \tan Q \) is \( 1 \).