Given: ∠AOB is a central angle and ∠ACB is a circumscribed angle.
Prove: △ACO ≅ △BCO
Circle O is shown. Line segments A O and B O are radii. Tangents C B and C B intersect at point C outside of the circle. A line is drawn to connect points C and O.
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 .

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Patterson16 Patterson16 · Answer 1 · 2021-04-29T17:19:12+02:00

Answer:

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

✔ all radii of the same circle are congruent.

We also know that AC ≅ BC since

✔ tangents to a circle that intersect are congruent.

Using the reflexive property, we see that

✔ side CO is congruent to side CO.

Therefore, we conclude that △ACO is congruent to △BCO by the

✔ SSS congruence theorem.

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AFOKE88 AFOKE88 · Answer 2 · 2022-03-23T20:22:35+01:00

The two column proof of the fact that △ACO is congruent to △BCO is; SSS Congruence Theorem

We are given that;

angle AOB is a central angle of circle O

angle ACB is a circumscribed angle of circle O.

Now, AO ≅ BO. The reason is because all radii of the same circle are congruent.

We are told that; AC ≅ BC. The reason is because tangents to a circle that intersect each other are congruent.

Using the reflexive property, we see that side CO is congruent to side CO.

Finally, we can conclude that △ACO is congruent to △BCO by the SSS congruence theorem.

Read more about the two column proof at; https://brainly.com/question/1788884