False
If radius of a circle intersects a chord then it bisects the chord only if radius is perpendicular to the chord.
true, because both distances of the chord are congruent
false
False. 1). The proposed equation y=mx suggests that the chord's right bisector has no y-intercept, i.e. passes through the origin. This is interesting, and appears plausible, and I'm willing to acknowledge that this aspect of it is true. But ... 2). If the slope of the chord is 'm', then the slope of its right bisector is not also 'm'. If it were, that would make the chord and its bisector parallel, which would be pretty silly. The slope of any line perpendicular to the chord, including its right bisector, has to be '-1/m'. The equation of the chord's right bisector is: Y = -X/m .
False
its false
False
False
If radius of a circle intersects a chord then it bisects the chord only if radius is perpendicular to the chord.
false
false
False because that would be its radius and the largest chord of a circle is its diameter.
true, because both distances of the chord are congruent
false
The perpendicular bisector of ANY chord of the circle goes through the center. Each side of a triangle mentioned would be a chord of the circle therefore it is true that the perpendicular bisectors of each side intersect at the center.
False