In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Sure, here is the step by step solution:

Step 1: In the given figure, join OQ, OP and OR.

Step 2: Since the tangent at a point to a circle is perpendicular to the radius through the point of contact, ∠ORQ = ∠ORP = 90°.

Step 3: Now, in triangle ORQ and ORP, we have ∠ORQ = ∠ORP.

Step 4: Also, ∠OQR = ∠OPR (Each 90°)

Step 5: Therefore, by the Angle-Angle (AA) criterion, we have ΔORQ ≅ ΔORP.

Step 6: Hence, OR = OP = OQ (Radii of the same circle)

Step 7: Now, in ΔPRQ, we have ∠PRQ = 120°.

Step 8: Also, ∠PQR = ∠PRQ = 120° (Angles subtended by the same arc at the circumference are equal)

Step 9: Therefore, ∠PQR + ∠PRQ + ∠RPQ = 180° (Angle sum property of a triangle)

Step 10: Substituting the values, we get 120° + 120° + ∠RPQ = 180°.

Step 11: Simplifying, we get ∠RPQ = -60°, which is not possible.

Step 12: Therefore, the given condition that ∠PRQ = 120° is not possible if OR = PR + RQ.

So, the statement "If ∠PRQ = 120°, then OR = PR + RQ" is not true.