Sure, here is the step by step proof:

Step 1: Draw the radius OA and OB to the points of tangency A and B.

Step 2: Since the radius is perpendicular to the tangent at the point of tangency, ∠OAP and ∠OBP are right angles.

Step 3: In right triangles OAP and OBP, we have OA = OB (radii of the same circle), AP = BP (tangents from an external point to a circle are equal), and ∠OAP = ∠OBP = 90°.

Step 4: By the RHS (Right angle-Hypotenuse-Side) criterion of congruence, we have ΔOAP ≅ ΔOBP.

Step 5: Therefore, ∠POA = ∠POB (corresponding parts of congruent triangles are equal).

Step 6: Also, OP = OP (common side).

Step 7: So, by the SAS (Side-Angle-Side) criterion of congruence, we have ΔPOA ≅ ΔPOB.

Step 8: Therefore, PA = PB (corresponding parts of congruent triangles are equal).

Step 9: Hence, OP is the right bisector of AB.

This completes the proof.

Question

Sure, here is the step by step proof:

Step 1: Draw the radius OA and OB to the points of tangency A and B.

Step 2: Since the radius is perpendicular to the tangent at the point of tangency, ∠OAP and ∠OBP are right angles.

Step 3: In right triangles OAP and OBP, we have OA = OB (radii of the same circle), AP = BP (tangents from an external point to a circle are equal), and ∠OAP = ∠OBP = 90°.

Step 4: By the RHS (Right angle-Hypotenuse-Side) criterion of congruence, we have ΔOAP ≅ ΔOBP.

Step 5: Therefore, ∠POA = ∠POB (corresponding parts of congruent triangles are equal).

Step 6: Also, OP = OP (common side).

Step 7: So, by the SAS (Side-Angle-Side) criterion of congruence, we have ΔPOA ≅ ΔPOB.

Step 8: Therefore, PA = PB (corresponding parts of congruent triangles are equal).

Step 9: Hence, OP is the right bisector of AB.

This completes the proof.

Knowee AI · Accepted Answer

Sure, here is the step by step proof:

Step 1: Draw the radius OA and OB to the points of tangency A and B.

Step 2: Since the radius is perpendicular to the tangent at the point of tangency, ∠OAP and ∠OBP are right angles.

Step 3: In right triangles OAP and OBP, we have OA = OB (radii of the same circle), AP = BP (tangents from an external point to a circle are equal), and ∠OAP = ∠OBP = 90°.

Step 4: By the RHS (Right angle-Hypotenuse-Side) criterion of congruence, we have ΔOAP ≅ ΔOBP.

Step 5: Therefore, ∠POA = ∠POB (corresponding parts of congruent triangles are equal).

Step 6: Also, OP = OP (common side).

Step 7: So, by the SAS (Side-Angle-Side) criterion of congruence, we have ΔPOA ≅ ΔPOB.

Step 8: Therefore, PA = PB (corresponding parts of congruent triangles are equal).

Step 9: Hence, OP is the right bisector of AB.

This completes the proof.

From a point P outside a circle with centre O, tangents PA and PB are drawn to the circle. Prove that OP is the right bisector of the segment AB.

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