1. **Problem statement:** Given circle (O) and point A outside (O), with tangents AB and AC to (O) at points B and C respectively. A secant ADE is drawn such that ray AD lies between rays AB and AO, with D between A and E. OK is perpendicular to DE at K (K lies on DE). 2. **Goal:** Prove that quadrilateral ABKO is cyclic. 3. **Key concepts:** - A quadrilateral is cyclic if and only if the opposite angles sum to 180°. - Tangents from a point to a circle are equal in length. - The radius to the point of tangency is perpendicular to the tangent line. 4. **Proof steps:** - Since AB and AC are tangents from A to circle (O), OA is the radius to the center O. - At point B, the radius OB is perpendicular to tangent AB, so \(\angle OBA = 90^\circ\). - Similarly, OK is perpendicular to DE by construction. - Consider quadrilateral ABKO: angles at B and K are right angles. - Since \(\angle OBA = 90^\circ\) and \(\angle OKB = 90^\circ\), their sum is 180°, so ABKO is cyclic by the converse of the cyclic quadrilateral theorem. **Final answer:** Quadrilateral ABKO is cyclic because it has two opposite right angles summing to 180°.