1. Copy a given line segment AB onto a given line L
Place the compass with point at one endpoint of
the line segment and pencil tip at the other, thus
setting the compass to the length of the line segment. Label one point A′ on the line L, place the
point of the compass at A′ and use it to mark off a
point B′ on the line. The segment A′B′ is congruent
to the segment AB.
2. Draw a line perpendicular to a given line L
through a given point P on L.
Set the compass at an arbitrary radius and, with
the tip of the compass placed on point P, mark off
two points A and B on line L equidistant from P.
Now draw two circles of the same radius, one with
center at A and the other with center at B. These
circles intersect at two positions X and Y. These
points X and Y are the same distance from each of
X and Y, and so, as the study of EQUIDISTANCE
proves, the line through X and Y (and P) is the perpendicular bisector of the line segment AB. In particular, it is a line through P perpendicular to L.
3. Construct the perpendicular bisector of a given line
segment AB.
The construction described in 2 above accomplishes
this feat.
4. Draw a line perpendicular to a given line L through
a given point P not on L.
Set the compass with its point at P and draw a
large circle that intersects the line L at two points
A and B. Now follow the procedure for 2.
5. Draw an equilateral triangle.
The points A, B, and X described in 2 are the vertices of an equilateral triangle.
6. Copy an arbitrary triangle to a different position
on the page.
Suppose the given triangle has vertices labeled A,
B, and C. Set the compass with point at A and tip
at B. Arbitrarily choose a point A′ elsewhere on the
page, and use the compass to draw a circle with
center A′ and radius equal to length of the segment
AB. Label an arbitrary point on this circle B′. Use
the compass to draw a second circle with center A′,
but this time with radius equal to the length of AC.
We must now select an appropriate point C′ on this
second circle. Draw a third circle with center B′ of
radius equal to the length of BC. Label a point of
intersection between the final two circles C′. Then
A′B′C′ is a congruent copy of the original triangle.
7. Copy a given angle to a different location on
the page.
Simply regard the angle as part of a triangle and
follow the instructions of part 6.
8. Construct a line parallel to a given line L through a
point P not on L.
Draw an arbitrary line through P that intersects the
line L. Copy the angle these two lines make at
position P and draw a third line through P at this
angle. This produces a diagram of a TRANSVERSAL
crossing a pair of lines possessing equal corresponding angles. By the converse of the PARALLEL
POSTULATE, the two lines are parallel.
9. Construct a line that divides a given angle precisely
in half.
Draw a circle of arbitrary radius with center at the
vertex of the angle. Suppose this circle intersects
the rays of the angles at positions A and B. Now
draw two circles of the same radius centered about
each of these two points. Let P be a point of intersection of the two circles. Then the line connecting
the vertex of the angle to P is an angle bisector.
(The SSS principle of similarity shows that the two
triangles produced in the construction are congruent, demonstrating then that the original angle is
indeed divided into two equal measures.)
10. Draw a perfect square.
Draw an arbitrary line segment. This will be the
first side of the square. Label its endpoints A and
B. Using part 2, construct a line through B perpendicular to the line segment. Use a circle centered
about B of radius equal in length to AB to find a
point C on this perpendicular line so that BC is the
same length as AC. This provides the second side
of the square. Repeat this procedure to construct
the remaining two sides of the square.
Not every geometric feat can be accomplished with
straightedge and compass alone. For example, although
it is possible to also construct a regular pentagon and a
96 constructible
Place the compass with point at one endpoint of
the line segment and pencil tip at the other, thus
setting the compass to the length of the line segment. Label one point A′ on the line L, place the
point of the compass at A′ and use it to mark off a
point B′ on the line. The segment A′B′ is congruent
to the segment AB.
2. Draw a line perpendicular to a given line L
through a given point P on L.
Set the compass at an arbitrary radius and, with
the tip of the compass placed on point P, mark off
two points A and B on line L equidistant from P.
Now draw two circles of the same radius, one with
center at A and the other with center at B. These
circles intersect at two positions X and Y. These
points X and Y are the same distance from each of
X and Y, and so, as the study of EQUIDISTANCE
proves, the line through X and Y (and P) is the perpendicular bisector of the line segment AB. In particular, it is a line through P perpendicular to L.
3. Construct the perpendicular bisector of a given line
segment AB.
The construction described in 2 above accomplishes
this feat.
4. Draw a line perpendicular to a given line L through
a given point P not on L.
Set the compass with its point at P and draw a
large circle that intersects the line L at two points
A and B. Now follow the procedure for 2.
5. Draw an equilateral triangle.
The points A, B, and X described in 2 are the vertices of an equilateral triangle.
6. Copy an arbitrary triangle to a different position
on the page.
Suppose the given triangle has vertices labeled A,
B, and C. Set the compass with point at A and tip
at B. Arbitrarily choose a point A′ elsewhere on the
page, and use the compass to draw a circle with
center A′ and radius equal to length of the segment
AB. Label an arbitrary point on this circle B′. Use
the compass to draw a second circle with center A′,
but this time with radius equal to the length of AC.
We must now select an appropriate point C′ on this
second circle. Draw a third circle with center B′ of
radius equal to the length of BC. Label a point of
intersection between the final two circles C′. Then
A′B′C′ is a congruent copy of the original triangle.
7. Copy a given angle to a different location on
the page.
Simply regard the angle as part of a triangle and
follow the instructions of part 6.
8. Construct a line parallel to a given line L through a
point P not on L.
Draw an arbitrary line through P that intersects the
line L. Copy the angle these two lines make at
position P and draw a third line through P at this
angle. This produces a diagram of a TRANSVERSAL
crossing a pair of lines possessing equal corresponding angles. By the converse of the PARALLEL
POSTULATE, the two lines are parallel.
9. Construct a line that divides a given angle precisely
in half.
Draw a circle of arbitrary radius with center at the
vertex of the angle. Suppose this circle intersects
the rays of the angles at positions A and B. Now
draw two circles of the same radius centered about
each of these two points. Let P be a point of intersection of the two circles. Then the line connecting
the vertex of the angle to P is an angle bisector.
(The SSS principle of similarity shows that the two
triangles produced in the construction are congruent, demonstrating then that the original angle is
indeed divided into two equal measures.)
10. Draw a perfect square.
Draw an arbitrary line segment. This will be the
first side of the square. Label its endpoints A and
B. Using part 2, construct a line through B perpendicular to the line segment. Use a circle centered
about B of radius equal in length to AB to find a
point C on this perpendicular line so that BC is the
same length as AC. This provides the second side
of the square. Repeat this procedure to construct
the remaining two sides of the square.
Not every geometric feat can be accomplished with
straightedge and compass alone. For example, although
it is possible to also construct a regular pentagon and a
96 constructible