Given a segment AB and a ray −−→CD, there is a point X on
−−→CD so that AB ∼= CX.
Proof. Let AB and −−→CD be given. Figure 1: Given starting point for
Theorem 28
Theorem 29 (Problem 18). Given two points A and B (see Figure 2, there is a third
point C not on
←→AB such that A, B, and C form an equilateral triangle.
Proof. Refer to Figure2.
Figure 2: Given starting point
and suggested construction for
Theorem 29
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Theorem 30 (Problem 21). Given ∠BAC and a ray −−→DE, there is a ray −−→DH on a given
side of line ←→DE so that ∠BAC ∼= ∠EDH.
Proof. Suppose we have ∠BAC and a ray −−→DE. Using a
compass, measure AB and copy this length along ray −−→DE
at point D. Let G be the point on −−→DE such thatAB ∼=
DE (such a point exists by Theorem 28).
Figure 3: Given starting point
and start of construction for Theorem 30
Theorem 31 (Problem 22). Every angle has a bisector.
Proof. Refer to Figure 4. As in Theorem 30, our construction relies on creating congruent triangles, and proving the triangles are congruent as a way to show the angles are congruent (and therefore that we have bisected
the angle). Let ∠BAC be given. Use a compass to create circle f with center A and radius AB. Let D be the
point of intersection of f with −→AC. Next, construct circle
g with center B and radius BD, and construct circle h
with center D and radius DB. Let E be one of the points
of intersection of h and g.
We will prove that 4AED ∼= 4AEB.
Figure 4: Angle bisector construction
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Theorem 32 (Problem 24). There is a line perpendicular to a given line through a given
point not on the line.
Proof. Refer to Figure 5. Let line ←→AB and point C not
on the line be given. Construct segment AC. If ∠BAC
is a right angle, then AC is the perpendicular and we
are done. If not, construct a circle with center at C and
radius AC. Let D be the other point of intersection of
the circle with ←→AB, and construct CD. Next, construct
. . .
Figure 5: Perpendicular construction
Theorem 33 (Problem 25). There is a line perpendicular to a given line through a given
point on the line.
Proof. Refer to Figure 6. Let line ←→AB and point C
on the line be given. Construct circle k with center C
and positive radius (any length will work), and let D and
E be the points of intersection of the line with k. At
point D, construct circle m with center D and radius
DE. Similarly, at point E, construct circle n with center
E and radius DE. Let F be a point of intersection of m
and n, and construct DF and EF.
We will prove that 4CF E ∼= 4CF D.
Figure 6: Second perpendicular
construction
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Theorem 34 (Problem 26). Every segment has a midpoint.
Proof. Refer to Figure 7. Let line ←→AD be given. Construct a circle with center at A and radius AD, and a
circle at D with radius AD, and let C and E be the
points of intersection of the two circles. Construct AC,
DC, AE, and DE. From here, . . .
Figure 7: Start of midpoint construction
Theorem 35. The base angles of an isosceles triangle are congruent angles.
Proof. Let isosceles 4ABC be given, with AB ∼= CB. By Theorem 34, AC has a midpoint,
which we will call D. Then . . .
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Sample Solution