# Solutions to Math Homework Δ

# Solutions to Math Homework 4

**Ch 3.2 no. 1.
**(a) A point A is said to lie between points A and B, denoted A − C − B, if

A,B,C are distinct and d(A,B) = d(A,C) + d(C,B).

(b) For two distinct points A,B a line segment consists of the points A,B

and of all points C on the unique line passing through A and B such that A−C−B.

The points A and B are called the endpoints of .

(c) If A,B,C are three distinct non-colinear points, then the angle∠BAC consists

of the points on rays and of all points D that lie between a point on

and a point on The rays are called the sides of ∠BAC and the

point A is called the vertex of ∠BAC.

(d) An angle is obtuse if it has measure more than 90° . An angle is acute if it

has measure less than 90° . An angle is right if it has measure 90° .

(e) Two angles are adjacent of they have a common side and their intersection

is equal to that side.

(f) Two angles are vertical if they have a common vertex , their intersection is

equal to that vertex and if the sides of each angle can be ordered in such a way

that the union of the first sides of these angles is a line and the union of the second

sides of these angles is a line.

(g) Two angles are supplementary if the sum of their measures is 180° . Two

angles are complementary if the sum of their measures is 90° .

(h) The midpoint of a line segment is a point on the segment that is equidistant

from the endpoints of the segment.

(i) A bisector of an angle ∠BAC is a ray that is contained in ∠BAC and

such that m∠BAD = m∠DAC = 1/2m∠BAC.

(j) Two lines l

_{1}, l

_{2}are said to be perpendicular if they intersect at a point A such

that for any point B on l

_{1}and any point C on l

_{2}such that B ≠ A and C ≠ A the

angle ∠BAC is right.

(k) If A,B,C are three non-colinear points then a triangle Δ ABC is the intersection

of the angles ∠ABC, ∠BAC and ∠CBA. The points A,B,C are called

vertices of ΔABC and the segments are called the sides of ΔABC.

(l) Let A

_{1},A

_{2},A

_{3}, . . .A

_{n}(where n ≥ 4) be n distinct points in the plane.

Suppose that:

(1) No three distinct points in the list A

_{1},A

_{2},A

_{3}, . . .A

_{n}are colinear; and

(2) For each i = 2, . . . , n − 2 the intersection of the triangles ΔA

_{1}A

_{i}A

_{i+1}and

A

_{1}A

_{i+1}A

_{i+2}is equal to the segment and

(3) The union of the triangles ΔA

_{1}A

_{2}A

_{3},ΔA

_{1}A

_{3}A

_{4}, . . .Δ,A

_{1}A

_{n-1}A

_{n}is convex.

Then a polygon P(A

_{1}, . . . ,A

_{n}) is the union of triangles

ΔA

_{1}A

_{2}A

_{3},ΔA

_{1}A

_{3}A

_{4}, . . . ,ΔA

_{1}A

_{n-1}A

_{n}.

The points A

_{1}, . . . ,A

_{n}are called the vertices of P(A

_{1}, . . . ,A

_{n}) and the segments

are called the sides of P(A

_{1}, . . . ,A

_{n}).

(m) The interior of an angle (triangle, polygon) consists of all those points of

the angle (triangle, polygon) that do not lie on its sides.

**Ch 3.2 no. 4
**We will show that the relation of angle congruence is an equivalence relation.

Recall that the angles ABC and ∠A′B′C′ are said to be congruent, denoted

ABC ∠A′B′C′, if m∠ABC = m∠A′B′C′.

(1)

**Reflexivity.**For any angle ∠ABC we have m∠ABC = m∠ABC and hence

∠ABC ∠ABC.

(2) S

**ymmetry.**Suppose ∠ABC ∠A′B′C′.

Then m∠ABC = m∠A′B′C′. Therefore m∠A′B′C′ = m∠ABC and hence

A'B'C' ∠ABC.

(3)

**Transitivity.**Suppose ABC ∠A'B'C' and A'B'C' ∠A"B"C"

Then m∠ABC = m∠A′B′C′ and m∠A′B′C′ = m∠A"B"C". Hence m∠ABC =

m∠A"B"C" and so ABC ∠A"B"C"

**Ch 3.2 no. 6.
**We need to prove that supplements and complements of congruent angles are

congruent. We will do that for complements.

Suppose α, β , are angles such that and let
β be a complementary angle of

α
and β' be a complementary angle of α'. We need to show that

Since we have mα = mα' . Since β is a complement of
α and β' is a

complement of α', we have mβ = 90° − mα and mβ' = 90° − mα' . Hence

Thus and as required.

**Ch 3.2 no. 7. **(Draw a picture for the argument below)

We need to prove that vertical angles are congruent.

Suppose angles ∠BAC and ∠B'AC' are vertical, so that
is a line l_{1}

and is a line l_{2}.

Since the angles ∠BAC and CAB' form a linear pair , SMSG Postulate 14 implies

that these angles are supplementary and so m∠BAC + m∠CAB' = 180° .

Similarly, the angles ∠CAB' and B'AC' form a linear pair and therefore by

SMSG Postulate 14 these angles are supplementary and m∠CAB' + m∠B'AC' =

180° .

Therefore

m∠C'AB' = 180° − m∠CAB' = 180° − (180° − m∠BAC) = m∠BAC

Thus m∠C'AB' = m∠BAC and so ∠C'AB'
∠BAC, as required.

**Ch 3.2 no. 9.** (Draw a picture for the argument below)

We need to prove that if a point is on a perpendicular bisector of a line
segment,

then it is equidistant from the endpoints of that segment.

Let l be a perpendicular bisector to a segment
and denote the midpoint of

by T.

Suppose C is a point on l.

If C = T then d(C,A) = d(C,B) =
and C is equidistant from A and

B. Suppose now that C ≠ T, so that A,B,C are not co- linear .

We have ∠CTA ∠CTB, since both angles are right. Moreover,
and

since d(A, T) = d(B, T).

Therefore by the SAS axiom (SMSG Postulate 15) ΔCTA
ΔCTB. This

implies that and so d(A,C) = d(B,C), as required.

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