geometry

need this back by February 08, 2022 at 12pm

Theorem

2

8 (Problem

1

7). 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

3

0 (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

Homework

2

Theorem

3

6. Congruent angles have congruent supplements.

Proof. Refer to Figure

1

. Let ∠BDA and ∠BDC be sup-
plementary, ∠EGH and ∠HGF be supplementary, and
further suppose we are given ∠BDA ∼= ∠EGH. We will
show that ∠BDC ∼= ∠HGF .

Figure 1: Diagram for congruent
supplements theorem

Corollary 37. Vertical angles are congruent.

Proof. Refer to Figure 2. Let

←→
AB and

←→
CD be distinct

lines and let X be a point that is on both lines. We wish
to show that ∠AXB ∼= ∠CXD.

Figure 2: Vertical angles

1

Theorem 38. An angle that is congruent to a right angle is also a right angle.

Proof. Refer to Figure 3. Let ∠BXA be given, with sup-
plementary ∠BXC, and let ∠EGH be given, with sup-
plementary ∠HGF , and suppose that ∠BXA is a right
angle, and ∠BXA ∼= ∠EGH. Then . . .

Figure 3: Weak right angle theo-
rem

Theorem 39. If two lines have a transversal which forms alternate interior angles that are
congruent, then the two lines are parallel.

Proof. Refer to Figure 4. This will be a proof by contra-

diction. Suppose that lines
←→
AB and

←→
CD are given, with

transversal
←→
EF, with G the point on

←→
AB and

←→
EF, and

H the point on
←→
CD and

←→
EF. Furthermore, suppose that

∠BGH ∼= ∠CHG. We will suppose that lines
←→
AB and

←→
CD are not parallel, and so intersect at point J, which

we may assume is on
−−→
AB.

Construct a circle with center H and radius the length of
GJ, and let K be the point of intersection of this circle

with
−−→
HC. Construct GK.

Figure 4: Alternatie interior an-
gles

2

Corollary 40. If two lines have a transversal which forms corresponding angles that are
congruent, then the two lines are parallel.

Proof.

Corollary 41. Given a line k and a point P not on k, there is a line m such that P is on
m and m is parallel to k.

Proof. Refer to Figure 5. Let
←→
AB be given, and point P

be given not on
←→
AB.

Figure 5: Parallel line construc-
tion

3