|
| Given: Arc GH of a circle |
| Required: To find the center of
the circle. |
| Procedure: |
|

|
Select any three points on arc GH. |
Label the points A, B, and C. |
Use the segment command under the construct menu
and construct segments AB and BC. |
Construct the perpendicular bisectors of these two segments.
This can be accomplished by following the basic perpendicular
bisector construction; (help, midpoint) (help, )
or by using the tools in Sketchpad which follows: |
Select segment AB. |
Use the midpoint command under the construct
menu to place a midpoint on AB. Label it D. |
Select the midpoint D and segment AB, use the perpendicular line
command under the construct menu and construct a perpendicular at
D. |
Select segment BC and repeat the above steps,
but label the midpoint E. |
Label the point of intersection of the two perpendicular bisectors
point F. |
|
F
is the center of the given circle/arc. |
|
| Proof: |
F is the intersection of the
bisectors of the sides of the triangle created by segments AB, BC, and
CA. |
Segment FA = FB = FC ( The
bisector of the sides of a triangle are concurrent in a point
equidistant from the vertices.) |
F is the center of the given circle/arc. |
|
| Sketch: Reposition
points A, B, or C on the arc of the circle in the sketch below.
Does the center remain the same? |
|