Proof of Equal Perpendiculars from Point M
To prove that the perpendiculars drawn from point M (on line AP) to lines AB and AC are equal, we can utilize the properties of angle bisectors and right triangles.

Key Concepts
- **Angle Bisector**: AP bisects angle BAC, meaning that ∠BAP = ∠PAC.
- **Perpendiculars**: Let MP be the perpendicular from M to AB and MQ be the perpendicular from M to AC.

Construction
1. **Draw Perpendiculars**: From point M, draw MP ⊥ AB and MQ ⊥ AC.
2. **Label Points**: Let the feet of the perpendiculars be P and Q respectively.

Triangles and Properties
- **Triangles MPB and MQC**: Since MP ⊥ AB and MQ ⊥ AC, triangles MPB and MQC are right triangles.
- **Angle Relationships**:
- In triangle MPB, ∠BAP = ∠PAC implies that the angles ∠MPB and ∠MQC are equal, as they both are supplementary to the same angles.

Using Similarity
- **Angle-Angle (AA) Criterion**:
- Since ∠BAP = ∠PAC and both triangles share the angle at M, triangles MPB and MQC are similar.
- **Corresponding Sides**: From the similarity of triangles, we have:
- MP/MQ = AB/AC

Conclusion
- **Equality of Perpendiculars**: Since AP bisects ∠BAC, we find that AB = AC. Hence, MP = MQ.
Thus, the perpendiculars from M to AB and AC are equal. This completes the proof.