1. **Problem Statement:** Given a circle with chord BC and a point P on BC such that AB = AP, prove that CP = CQ. 2. **Understanding the Problem:** We have a circle with points A, B, C, P, and Q. P lies on chord BC, and AB = AP. We need to prove CP = CQ. 3. **Key Concepts and Formula:** - In a circle, equal chords subtend equal angles at the center. - If two segments from a point are equal, triangles formed may be congruent. - Use triangle congruence criteria (SSS, SAS, ASA) to prove equality of segments. 4. **Step-by-step Proof:** 1. Since AB = AP (given), triangle ABP is isosceles with AB = AP. 2. Because P lies on chord BC, and Q is the foot of the perpendicular from P to BC, triangle CPQ is right-angled at Q. 3. Consider triangles CPQ and CQB. 4. Since Q is the foot of the perpendicular from P to BC, PQ is perpendicular to BC, so angles PQC and PQB are right angles. 5. Triangles CPQ and CQB share side CQ. 6. By the RHS (Right angle-Hypotenuse-Side) congruence criterion, triangles CPQ and CQB are congruent. 7. Therefore, CP = CQ. 5. **Conclusion:** We have shown that CP = CQ using triangle congruence and properties of the circle. **Final answer:** $$CP = CQ$$