Statement: In a circle with center O, if two tangents PQ and PR are drawn from an external point P such that ∠QPR = 120°, then 2PQ = PO.

Proof:

1. Construction:
Let's construct the diagram as described in the problem statement.

1. Draw a circle with center O.
2. Choose an external point P.
3. Draw two tangents PQ and PR from point P to the circle.
4. Join OP.

2. Proof by contradiction:
To prove the statement, we will assume the opposite and show that it leads to a contradiction.

Assume that 2PQ ≠ PO.

3. Triangle OQP:
Consider triangle OQP.

1. Since PQ is a tangent to the circle, it is perpendicular to the radius drawn to the point of tangency Q. Therefore, ∠OQP = 90°.
2. Similarly, since PR is a tangent to the circle, ∠OPR = 90°.

4. ∠QPR = 120°:
Given that ∠QPR = 120°.

5. ∠OPR = 60°:
Since ∠OPR = 90° and ∠QPR = 120°, we can conclude that ∠OPR = 180° - ∠QPR = 60°.

6. ∠OQR = 30°:
Since ∠OPQ = ∠OPR - ∠QPR = 60° - 120° = -60° (interior angles of a triangle), we can conclude that ∠OQR = 180° - ∠OPQ = 180° - (-60°) = 240°.

7. Triangle OQR:
Consider triangle OQR.

1. Since ∠OQR = 30°, ∠OQP = ∠QOP = (180° - ∠OQR)/2 = (180° - 240°)/2 = -60°/2 = -30°/2 = -15°.
2. Therefore, ∠OQP + ∠OPQ + ∠QPO = -15° + 90° + ∠QPO = 75° + ∠QPO.

8. ∠QPO ≠ 0°:
Since PQ and PR are tangents from point P, ∠QPO ≠ 0°.

9. Contradiction:
From step 7, we know that ∠OQP + ∠OPQ + ∠QPO = 75° + ∠QPO. But from step 8, we know that ∠QPO ≠ 0°. Therefore, ∠OQP + ∠OPQ + ∠QPO ≠ 180°.

This contradicts the fact that the sum of angles in a triangle is always 180°.

10. Conclusion:
Our assumption that 2PQ ≠ PO leads to a contradiction. Therefore, the statement 2PQ = PO must be true.

Hence, it is proved that if two tangents PQ and PR are drawn from an external point P to a circle with center O,