Question

given: ∠aob is a central angle and ∠acb is a circumscribed angle. prove: △aco ≅ △bco we are given that angle aob is a central angle of circle o and that angle acb is a circumscribed angle of circle o. we see that (overline{ao} cong overline{bo}) because all radii of the same circle are congruent. we also know that (overline{ac} cong overline{bc}) since tangents to a circle that intersect are congruent. using the reflexive property, we see that dropdown with options: side ca is congruent to side cb, side ao is congruent to side bo, side co is congruent to side co, side ab is congruent to side co o is congruent to dropdown.

Explanation:

The reflexive property states that a segment is congruent to itself. In triangles \( \triangle ACO \) and \( \triangle BCO \), the side \( CO \) is common to both triangles. So, by the reflexive property, \( \overline{CO} \cong \overline{CO} \). Among the given options, "side \( CO \) is congruent to side \( CO \)" is the correct application of the reflexive property here.

Answer:

side \( CO \) is congruent to side \( CO \)