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Remember
-- use your compass and straight edge only! |
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Bisect
- cut into two congruent (equal) pieces.
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Bisect
a line segment
(Also
know as Construct a Perpendicular Bisector of a segment) |
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Given:
(Line segment)

Task: Bisect
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Directions:
(It may be
advantageous to instruct students in the use of the "large arc method" because it creates a
"crayfish" looking creature which students easily remember and
which reinforces the circle concept needed in the explanation of the
construction.)
Explanation
of construction:
To understand the explanation you will need to label the point of
intersection of the arcs above segment
as D and below segment
as
E. Draw segments
,
,
and
. All four of these segments
are of the same length since they are radii of two congruent circles. More specifically, DA = DB and EA = EB.
Now, remember a locus theorem: The locus of points equidistant from
two points, is the perpendicular bisector of the line segment determined
by the two points. Hence,
is the perpendicular
bisector of .
The fact that the
bisector is also perpendicular to the segment is
actually MORE than we needed for a simple "bisect" construction.
Isn't this great! Free stuff!!!

Directions:
1. Place the point of the compass on the vertex of
(point A).
2. Stretch the compass to any length so long as it stays ON the
angle.
3. Swing an arc so the pencil crosses both sides of
. This will create two intersection points with the sides
(rays) of
the angle.
4. Place the compass point on one of these
new intersection points on the sides of
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If needed, stretch your compass to a
sufficient length to place your pencil well into the interior of the
angle. Stay between the sides (rays) of the angle. Place an
arc in this interior - you do not need to cross the sides of the angle.
5. Without changing the width of the compass, place the point of the
compass on the other intersection point on the side of the angle and make
the same arc. Your two small arcs in the interior of the angle
should be crossing.
6. Connect the point where the two small arcs cross to the vertex
A
of the angle.You have now created
two new angles that are of equal measure (and are each 1/2 the measure of
.)
Explanation of
construction:
To
understand the explanation, some additional labeling will be needed.
Label the point where the arc crosses side
as
D. Label the point
where the arc crosses side
as E. And label the intersection of
the two small arcs in the interior as F. Draw segments
and
.
By the construction, AD = AE (radii of same circle) and DF = EF (arcs of
equal length). Of course AF = AF. All of these sets of equal
length segments are also congruent. We have congruent triangles by SSS.
Since the triangles are congruent, any of their leftover corresponding
parts are congruent which makes
equal (or congruent) to
.

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