GEOMETRY

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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