Question

use the information to complete the problem.
in the figure, line ( t ) is tangent to circle ( a ) at point ( b ). point ( c ) lies on line ( t ) outside of the circle.
prove, by contradiction, that the radius, ( overline{ab} ), of circle ( a ) is perpendicular to the tangent line, ( t ). move options to the lines to complete the proof.
assume that point ( c ) lies on line ( t ) such that ( c
eq b ) and ( overline{ac} perp t ).
by the definition of perpendicular lines, ( ) must be a right angle. then, ( ) must be acute because the angles in a triangle add to ( 180^{circ} ). this means ( ) < ( ) because larger sides are opposite larger angles in a triangle. however, by the definition of a tangent line, point ( ) lies on the circle and point ( ) lies outside the circle. this means that the length of ( ) < ( ).
this proves, by contradiction, that ( overline{ab} perp t ).

Explanation:

Step1: Analyze triangle properties

In a triangle, if \(AC\perp t\) (assumed as in the proof structure), \(\angle ABC\) is a right - angle (by the definition of perpendicular lines). Then, using the triangle angle - sum property (\(\angle ABC+\angle ACB+\angle BAC = 180^{\circ}\)), if \(\angle ABC = 90^{\circ}\), and considering the side - angle relationships.

Step2: Use side - length and angle relationships

Since point \(C\) is outside the circle, \(AC>AB\) (a radius \(AB\) and a line segment \(AC\) from the center \(A\) to a point \(C\) outside the circle). In \(\triangle ABC\), by the side - angle relationship (larger side is opposite larger angle), \(\angle ACB<\angle ABC\)

Answer:

\(\angle ABC\): right angle; \(\angle ACB\): acute angle; \(AB\): radius; \(AC\): line segment from center to outside point; \(A\): center; \(B\): point of tangency; \(C\): outside point. So the order is \(\angle ABC\) (right angle), \(\angle ACB\) (acute angle), \(AB\) (radius), \(AC\) (\(AC > AB\)), \(A\) (center), \(B\) (point of tangency), \(C\) (outside point)