Yes, any diameter which is perpendicular to a chord bisects said chord.
This can be proved most easily with a picture, but is proved using a congruent triangle proof. Both triangles include the points at the center of the circle and the intersection of the diameter and chord. The other points should be the endpoints of the chord. They are congruent by hypotenuse leg; it was given that they are right triangle by the "perpendicular", the "leg" is the segment between the center of the circle and the intersection, and it is equal in both triangles because it is the same segment in both triangles. The hypotenuses are equal because both are radii of the circle.
Because the triangles are congruent, their sides must be so the two halves of the chord are congruent, and therefore the chord is bisected by the diameter.
Only when one of them is the circle's diameter which is the circle's largest chord.
The radius of the circle that is perpendicular to a chord intersects the chord at its midpoint, so it is said to bisect the chord.
because the chord can be determine by the diameter and the diameter can be determine by the chord.
Easy if you know how to draw a perpendicular and to bisect angles. Draw a perpendicular = 90 degrees. Bisect it = 45 degrees Bisect that = 22.5 degrees. Not sure how you can do it without that knowledge.
A diameter is a chord, because a chord always goes from one point to another in a circle, and a diameter does too.
yes
Only when one of them is the circle's diameter which is the circle's largest chord.
no.
The radius of the circle that is perpendicular to a chord intersects the chord at its midpoint, so it is said to bisect the chord.
. . the chord.
I get 9
because the chord can be determine by the diameter and the diameter can be determine by the chord.
The longest chord of a circle is its diameter
Bisects
Perpendicular.
No, it cannot.
Yes. it is possible to bisect a segment with a perpendicular segment. Follow the link to learn how to do it: http://www.mathopenref.com/constbisectline.html