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 ao≅bo because we also know that ac≅bc since using the reflexive property, we see that therefore, we conclude that △aco is congruent to △bco by the sss congruency theorem asa congruency theorem hl congruency theorem aas congruency theorem

Explanation:

Step1: Identify equal - radius sides

$AO$ and $BO$ are radii of the same circle, so $AO = BO$.

Step2: Identify equal - tangent sides

Tangents drawn from an external point to a circle are equal. Since $AC$ and $BC$ are tangents from point $C$ to circle $O$, $AC = BC$.

Step3: Identify the common side

$CO$ is common to both $\triangle ACO$ and $\triangle BCO$, so $CO=CO$ by the reflexive property.

Step4: Apply congruence theorem

Since $AO = BO$, $AC = BC$ and $CO = CO$, $\triangle ACO\cong\triangle BCO$ by the SSS (Side - Side - Side) congruency theorem.

Answer:

$AO$ and $BO$ are radii of the same circle; Tangents drawn from an external point to a circle are equal; $CO = CO$; SSS congruency theorem