Geometry need this back by February 08, 2022 at 12pm Theorem 28 (Problem 17). Given a segment AB and a ray −−→ CD, there is a point X on −−→ CD so that AB

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need this back by February 08, 2022 at 12pm

Theorem 28 (Problem 17). 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

1

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 The-
orem 30

Theorem 31 (Problem 22). Every angle has a bisector.

Proof. Refer to Figure 4. As in Theorem 30, our con-
struction relies on creating congruent triangles, and prov-
ing the triangles are congruent as a way to show the an-
gles are congruent (and therefore that we have bisected
the angle). Let ∠BAC be given. Use a compass to cre-
ate 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 con-
struction

2

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 con-
struction

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 4CFE ∼= 4CFD.

Figure 6: Second perpendicular
construction

3

Theorem 34 (Problem 26). Every segment has a midpoint.

Proof. Refer to Figure 7. Let line
←→
AD be given. Con-

struct 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 con-
struction

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 . . .

4

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