In a `Delta ABC, angle B = 2 angle C ` Angle bisector AD and BE intersect of O. AB = CD find `angle ABC `

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Angles
Let \( \angle C = y \). According to the problem, \( \angle B = 2\angle C \), so we can express \( \angle B \) as:
\[ \angle B = 2y \]
Let \( \angle A = x \).

### Step 2: Use the Triangle Angle Sum Property
In triangle \( ABC \), the sum of the angles is equal to \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the expressions for \( \angle A \) and \( \angle B \):
\[ x + 2y + y = 180^\circ \]
This simplifies to:
\[ x + 3y = 180^\circ \]

### Step 3: Use the Angle Bisector Theorem
Since \( AD \) and \( BE \) are angle bisectors, we know that:
\[ \angle DAC = \angle ACD \]
This implies that the angles opposite to equal sides are equal. Given that \( AB = CD \), we can state:
\[ x = y \]

### Step 4: Substitute \( x \) in the Angle Sum Equation
Now we substitute \( x \) with \( y \) in the equation \( x + 3y = 180^\circ \):
\[ y + 3y = 180^\circ \]
This simplifies to:
\[ 4y = 180^\circ \]
Thus, we find:
\[ y = \frac{180^\circ}{4} = 45^\circ \]

### Step 5: Find \( \angle B \)
Now we can find \( \angle B \):
\[ \angle B = 2y = 2 \times 45^\circ = 90^\circ \]

### Step 6: Find \( \angle A \)
Finally, we can find \( \angle A \):
\[ \angle A = 180^\circ - \angle B - \angle C = 180^\circ - 90^\circ - 45^\circ = 45^\circ \]

### Conclusion
Thus, the angles in triangle \( ABC \) are:
- \( \angle A = 45^\circ \)
- \( \angle B = 90^\circ \)
- \( \angle C = 45^\circ \)

The required angle \( \angle ABC \) is:
\[ \angle ABC = 90^\circ \]