In Figure, line `l` is the bisector of angle `A` and `B` is any point on `l` . `B P` and `B Q` are perpendiculars from `B` to the arms of `Adot` Show that : `A P B~=A Q B` `B P=B Q` or `B` is equidistant from the arms of `/_A` . Figure

Explanation

To solve the problem, we need to show that triangles \( APB \) and \( AQB \) are congruent, and from that, we will conclude that \( BP = BQ \), which indicates that point \( B \) is equidistant from the arms of angle \( A \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Line \( l \) is the bisector of angle \( A \).
- Point \( B \) lies on line \( l \).
- \( BP \) and \( BQ \) are perpendiculars from point \( B \) to the arms of angle \( A \).
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