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}congoverline{bo}) because we also know that (overline{ac}congoverline{bc}) since using the reflexive property, we see that therefore, we conclude that △aco is congruent to △bco by the

Explanation:

Step1: Radii of a circle are equal

$\overline{AO}$ and $\overline{BO}$ are radii of circle $O$, so $\overline{AO}\cong\overline{BO}$.

Step2: Tangent - segment theorem

Tangent segments from an external point to a circle are congruent. Since $A$ and $B$ are points of tangency and $C$ is the external point, $\overline{AC}\cong\overline{BC}$.

Step3: Reflexive property

$\overline{CO}\cong\overline{CO}$.

Step4: SSS congruence criterion

Since $\overline{AO}\cong\overline{BO}$, $\overline{AC}\cong\overline{BC}$ and $\overline{CO}\cong\overline{CO}$, by the Side - Side - Side (SSS) congruence criterion, $\triangle ACO\cong\triangle BCO$.

Answer:

$\triangle ACO\cong\triangle BCO$ by SSS (Side - Side - Side) congruence criterion.