ABCD is a square. BEC is an equilateral triangle inside the square. If CE & BD intersects at O. Find `angleBOC=?`

Answer

To find the angle \( \angle BOC \) where \( O \) is the intersection of lines \( CE \) and \( BD \) in the given configuration, we can follow these steps:

### Step 1: Understand the Configuration
- We have a square \( ABCD \) with vertices \( A, B, C, D \).
- Inside this square, there is an equilateral triangle \( BEC \) where \( B \) and \( C \) are two vertices of the square and \( E \) is a point inside the square.

### Step 2: Identify Angles in the Equilateral Triangle
- In an equilateral triangle, all angles are equal to \( 60^\circ \).
- Therefore, \( \angle BEC = 60^\circ \).

### Step 3: Determine Angles in the Square
- Since \( ABCD \) is a square, all angles are \( 90^\circ \).
- Thus, \( \angle ABC = 90^\circ \) and \( \angle BCD = 90^\circ \).

### Step 4: Analyze the Intersection of Lines
- The lines \( BD \) and \( CE \) intersect at point \( O \).
- We need to find \( \angle BOC \).

### Step 5: Use the Angle Properties
- We know that \( \angle BEC = 60^\circ \) (from the equilateral triangle).
- The angle \( \angle BOC \) can be found using the property of angles around point \( O \):
\[ \angle BOC + \angle BEC = 180^\circ \]
This is because \( \angle BEC \) and \( \angle BOC \) are supplementary angles.

### Step 6: Solve for \( \angle BOC \)
- Substitute \( \angle BEC \) into the equation:
\[ \angle BOC + 60^\circ = 180^\circ \]
- Rearranging gives:
\[ \angle BOC = 180^\circ - 60^\circ = 120^\circ \]

### Final Answer
Thus, the measure of \( \angle BOC \) is \( 120^\circ \).

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