In trapezium ABCD, AB is parallel to DC. P and Q are the mid-points of AD and BC respectively. BP product meets CD produced at point E. Prove that :
Point P bisects BE,

Answer

To prove that point P bisects BE in trapezium ABCD, where AB is parallel to DC, and P and Q are the midpoints of AD and BC respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Trapezium**:
- Draw trapezium ABCD with AB || DC.
- Mark points P and Q as the midpoints of sides AD and BC respectively.

2. **Extend Line CD**:
- Extend line CD to meet the line BP at point E.

3. **Identify Angles**:
- Since AB is parallel to DC, we know that angle ABP is equal to angle DPE (alternate interior angles).
- Similarly, angle APB is equal to angle PED (also alternate interior angles).

4. **Triangles Involved**:
- Consider triangles APB and DPE.
- We have:
- AP = PD (since P is the midpoint of AD)
- Angle APB = Angle DPE (as established above)
- Angle ABP = Angle PED (as established above)

5. **Apply the ASA Criterion**:
- Since we have two angles and the included side equal in both triangles, we can conclude that triangle APB is congruent to triangle DPE by the Angle-Side-Angle (ASA) criterion.

6. **Corresponding Parts of Congruent Triangles**:
- By the property of congruent triangles, we know that corresponding parts are equal.
- Therefore, PB = PE.

7. **Conclusion**:
- Since PB = PE, point P bisects line segment BE.
- Thus, we have proved that point P bisects BE.