Question

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

Answer 1

Answer: We can fill all the blanks with help of below explanation.

Explanation:

Given: $\angle AOB$ is a central angle and $\angle ACB$ is a circumscribed angle.  (where, AC and CB are two tangent lines from an external point C.)

Prove: $\triangle ACO\cong\triangle BCO$

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$AO\cong BO$ because both OA and OB make the radius of the circle O. (Since,  A and B are points on the circumference of circle O So, they are equally far from the center O

We also know that $AC\cong BC$ since both are the tangent line from a common point C So, they must be equal.(By the property of tangent line)

Using the reflexive property, we see that $OC\cong OC$

Therefore, we conclude that $\triangle ACO\cong\triangle BCO$ by the SSS postulate.

Answer 2

This is the Answer to the question