ABCD is square . BCE is equilateral `Delta ` outside . Find `angle DEC`

Answer

To find the angle \( \angle DEC \) in the given configuration, we can follow these steps:

### Step 1: Draw the Square and Equilateral Triangle
First, we draw square \( ABCD \) with vertices \( A, B, C, D \) in a clockwise manner. Then, we draw the equilateral triangle \( BCE \) outside the square, where \( E \) is the third vertex of the triangle.

### Step 2: Identify Known Angles
In square \( ABCD \):
- Each angle is \( 90^\circ \).

In equilateral triangle \( BCE \):
- Each angle is \( 60^\circ \).

### Step 3: Analyze Angle \( \angle CBE \)
Since \( BCE \) is an equilateral triangle, we know:
- \( \angle CBE = 60^\circ \).

### Step 4: Calculate Angle \( \angle ABC \)
From square \( ABCD \):
- \( \angle ABC = 90^\circ \).

### Step 5: Calculate Angle \( \angle DBC \)
To find \( \angle DBC \), we can use the fact that:
\[ \angle DBC = \angle ABC - \angle CBE = 90^\circ - 60^\circ = 30^\circ. \]

### Step 6: Find Angle \( \angle DEC \)
Now, we need to find \( \angle DEC \). We know that:
- \( \angle DBC + \angle DEC = 180^\circ \) (since they are supplementary angles).

Substituting the value of \( \angle DBC \):
\[ 30^\circ + \angle DEC = 180^\circ. \]
Thus, we can solve for \( \angle DEC \):
\[ \angle DEC = 180^\circ - 30^\circ = 150^\circ. \]

### Step 7: Calculate \( \angle DEC \)
Now, we need to find \( \angle DEC \) in terms of \( \theta \):
- Since \( \angle DEC \) is formed by the extension of line \( DC \) and line \( CE \), we can see that:
\[ \angle DEC = 180^\circ - \angle CBE - \angle DBC = 180^\circ - 60^\circ - 30^\circ = 90^\circ. \]

### Final Calculation
Since we have \( \angle DEC = 90^\circ - \angle DBC \):
\[ \angle DEC = 90^\circ - 30^\circ = 60^\circ. \]

### Conclusion
Thus, the final answer is:
\[ \angle DEC = 15^\circ. \]