The locus of a point on rhombus ABCD, so that it is equidistant from
AB and BC

Answer

To find the locus of a point on rhombus ABCD that is equidistant from sides AB and BC, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Rhombus**: Start by sketching the rhombus ABCD. Label the vertices A, B, C, and D in a clockwise manner.

2. **Draw the Angle Bisector**: Draw the angle bisector of angle A (∠A). This line will divide the angle A into two equal angles. Let this bisector intersect side BC at point E.

3. **Join Diagonal BD**: Draw the diagonal BD of the rhombus. This line will help in finding the intersection point with the angle bisector.

4. **Find Intersection Point O**: Identify the point O where the diagonal BD intersects the angle bisector AE.

5. **Draw Perpendiculars**: From point O, draw a perpendicular line OL to side AB and another perpendicular line OM to side AD.

6. **Analyze Triangles AOL and AOM**: Now, consider triangles AOL and AOM. We will prove that these two triangles are congruent.

7. **Congruence of Triangles**:
- Angle OAL = Angle OAM (because AE is the angle bisector).
- Angle OLA = Angle OMA = 90° (because OL and OM are perpendiculars).
- OA = OA (common side).
- By the Angle-Angle-Side (AAS) postulate, triangles AOL and AOM are congruent.

8. **Conclusion**: Since triangles AOL and AOM are congruent, the corresponding parts of these triangles are equal. Thus, OL = OM, which means that point O is equidistant from sides AB and AD.

Therefore, the locus of points that are equidistant from sides AB and BC is the angle bisector of angle A, which is the line AE.