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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amyb1216 amyb1216 · Answer 1 · 2021-04-29T16:14:26+02:00

Answer:

all radii of the same circle are congruent

tangents to a circle that intersect are congruent

side CO is congruent to side CO

SSS congruency theorem

Step-by-step explanation:

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jainveenamrata jainveenamrata · Answer 2 · 2022-08-04T10:37:35+02:00

The triangle △ACO ≅ △BCO are congruent to each other by SAS postulate.

Given: ∠AOB is a central angle and ∠ACB is a circumscribed angle.

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

We know that the radius and the tangent are perpendicular to each other.

Then the angle ∠CAO and angle ∠CBO are of 90°.

We know that the tangent of the circle drawn from the point outside the circle will be 2, and these tangent are of equal length.

In triangle △ACO ≅ △BCO, we have

CO = CO (Common side)

∠CAO = ∠CBO = 90°

OA = OB (Radius)

Thus, the triangle △ACO ≅ △BCO are congruent to each other by SAS postulate.

More about the triangle link is given below.

https://brainly.com/question/25813512

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