English | Español

Try our Free Online Math Solver!

Online Math Solver

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.


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, l2 are said to be perpendicular if they intersect at a point A such
that for any point B on l1 and any point C on l2 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 A1,A2,A3, . . .An (where n ≥ 4) be n distinct points in the plane.
Suppose that:
(1) No three distinct points in the list A1,A2,A3, . . .An are colinear; and
(2) For each i = 2, . . . , n − 2 the intersection of the triangles ΔA1AiAi+1 and
A1Ai+1Ai+2 is equal to the segment and
(3) The union of the triangles ΔA1A2A3,ΔA1A3A4, . . .Δ,A1An-1An is convex.
Then a polygon P(A1, . . . ,An) is the union of triangles
ΔA1A2A3,ΔA1A3A4, . . . ,ΔA1An-1An.
The points A1, . . . ,An are called the vertices of P(A1, . . . ,An) and the segments
are called the sides of P(A1, . . . ,An).
(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) Symmetry. 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 l1
and is a line l2.

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.

Prev Next