In a `Delta ABC` .G is centroid AG = BC find angle BGC

Answer

To solve the problem, we need to find the angle \( BGC \) in triangle \( ABC \) where \( G \) is the centroid and \( AG = BC \).

### Step-by-Step Solution:

1. **Draw Triangle ABC**:
Start by sketching triangle \( ABC \) with points \( A \), \( B \), and \( C \).

2. **Identify the Centroid G**:
The centroid \( G \) of triangle \( ABC \) is the point where the three medians intersect. A median is a line segment from a vertex to the midpoint of the opposite side.

3. **Label the Sides**:
Given that \( AG = BC \), we can denote \( BC = 2X \) for some value \( X \). Therefore, \( AG = 2X \).

4. **Determine the Segments**:
Since \( G \) is the centroid, it divides the median \( AG \) in the ratio \( 2:1 \). This means:
- \( AG = 2X \)
- \( GD = X \) (where \( D \) is the midpoint of \( BC \))

5. **Establish Angles**:
In triangle \( BGC \), since \( BG = GC \) (as \( G \) is the centroid and divides \( BC \) into two equal parts), we can denote the angles at \( B \) and \( C \) as \( \alpha \) and \( \beta \) respectively.

6. **Use the Isosceles Triangle Property**:
Since \( BG = GC \), angles \( \angle BGC \) and \( \angle GBC \) are equal. Thus, we can say:
\[ \angle BGC = \angle GBC = \alpha \]
and \( \angle BCG = \beta \).

7. **Apply the Triangle Sum Theorem**:
The sum of angles in triangle \( BGC \) is \( 180^\circ \):
\[ \alpha + \alpha + \beta = 180^\circ \]
Simplifying gives:
\[ 2\alpha + \beta = 180^\circ \]

8. **Relate Angles**:
Since \( AG = BC \) and we have established the segments, we can conclude that \( \alpha + \beta = 90^\circ \) (from the properties of the triangle).

9. **Final Calculation**:
Substituting \( \beta = 90^\circ - \alpha \) into the equation \( 2\alpha + \beta = 180^\circ \):
\[ 2\alpha + (90^\circ - \alpha) = 180^\circ \]
This simplifies to:
\[ \alpha + 90^\circ = 180^\circ \]
Therefore,
\[ \alpha = 90^\circ - \beta \]
Thus, \( \angle BGC = 90^\circ \).

### Conclusion:
The angle \( BGC \) is \( 90^\circ \).