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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

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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 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

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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 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

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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 . . .

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