Download The Area Concept , Similarity in Triangles, and Applications of the

The Area Concept , Similarity in Triangles,
and Applications of the Basic Proportionality Theorem
Section 4.3: The Area Concept (SMSG Postulates 17 – 20)
The three properties which are necessary for all “area measure assignments”:
1) Every polygon is assigned a unique number called its “area”,
2) Congruent triangles (polygons) are assigned the same “area number”,
3) If a polygon P is the union of two regions R1 and R2 (which intersect
only in a finite number of points and line segments), then
Area of P = (Area of R1) + (Area of R2)
Postulates 17, 18 and 19 ensure that the Euclidean Area Assignment meets requirements
1, 2 and 3, respectively.
Postulate 20: The area of a rectangle is the product: Area = Base × Altitude.
Theorem 4.3.1 (Proposition 35 in Book I of Euclid’s Elements):
Two parallelograms that share a common base and whose sides opposite this base are
contained in the same line are equal in area.
Proof:
Consider parallelograms ABCD and
EBCF that share the common base
and
whose opposite sides are contained in the same
line
.
A
By the Converse of the Alternate Interior Angle
Theorem, the angles indicated as congruent are
congruent.
D
E
F
T2
G
T1
B
C
Since the opposite sides of a parallelogram are congruent, the segments indicated as
congruent are congruent. Now, AE = AD + DE = EF + DE = DE + EF = DF .
Thus,

and so, Δ ABE  Δ DCF by ASA .
Since two congruent figures have equal area measure,
(Area of ΔABE) = (Area of ΔDCF) .
Let Δ T1 = Δ BCG and
Δ T2 = Δ DEG .
Area of ABCD = (Area of ΔABE) + (Area of T1) – (Area of T2)
and
Area of EBCF = (Area of ΔDCF) + (Area of T1) – (Area of T2) .
By substitution,
Area of ABCD = (Area of ΔDCF) + (Area of T1) – (Area of T2) = Area of EBCF . QED
2
The Height of Triangles and Parallelograms (Relative to a chosen base)
Definition: Given a triangle, choose one side to be called the base of the triangle.
D
C
h
h
A
B F1
C
A
F2
h
B
F3
The height h of the triangle (relative to the chosen base) is the length of the segment
drawn perpendicular to the base from the vertex not on the base. (See below.)
Given a parallelogram, choose one side to be called the base of the parallelogram.
The height h of the parallelogram (relative to the chosen base) is the length of the
segment drawn perpendicular to the base from any point on side opposite the base. (See
above.)
The Altitude of a Triangle (Relative to one of the sides)
Definition: The term “altitude of a triangle” has 3 meanings depending on the context.
Given Δ ABC, choose one side, say AB , with vertex C opposite the chosen side.
1. The line through C and perpendicular to the line containing AB is the altitude of the
triangle drawn from C to the opposite side AB .
2. The line segment drawn from C to the foot of the perpendicular line, drawn from C to
the line containing AB , is the altitude of the triangle drawn from C to the opposite side AB .
3. The length of the line segment drawn from C to the foot of the perpendicular line,
drawn from C to the line containing AB , is the altitude of the triangle relative to side AB .
In this sense, the altitude and the height of a triangle are the same.
There are three altitudes of a triangle, each relative to one the three sides of the triangle.
3
Theorem 4.3.2, The "Parallelogram Area Formula" Theorem:
The area of a parallelogram is the product of its height and the length of its base.
Proof: Theorem is true for a rectangle (SMSG Postulate 20).
D
A
C
E
B
F
These two parallelograms have the same height.
They share the same base (Segment CD) and the sides opposite the base lie along the
same line.
By the previous theorem, they have the same area, which for the rectangle is:
AREA = BASE × HEIGHT .
Theorem 4.3.3: The area of a Right Triangle is: AREA = ½ (Base) (Height).
C
A'
Δ ABC  Δ A’CB by SSS
Area(Δ ABC) = ½ Area(‫ٱ‬ABA’C)
= ½ AB × AC
A
B
= ½ (Base) (Height)
4
Theorem 4.3.4, The "Triangle Area Formula" Theorem:
The area of a Triangle is: AREA = ½ (Base) (Height) .
Proof: Every triangle has at least two acute angles. Assume that there are acute
angles at A and B. Drop a perpendicular segment from C to D on side AB.
C
Area(Δ ABC) =
Area(Δ ADC) + Area(Δ BDC) =
½ AD×CD + ½ DB×CD =
½ (AD + DB) × CD =
A
D
B
½ (Base AB) (Height CD)
Section 4.4: The Similarity Relation
Definition:
Two triangles Δ ABC and Δ A’B’C’ (or two n-gons …) are similar
( write Δ ABC ~ Δ A’B’C’ )
if there is a one-to-one correspondence between the vertices
A ↔ A’ , B ↔ B’ , C ↔ C’
such that:
1) Corresponding angles are congruent, and
2) Any two pairs of corresponding sides are in the same
proportion k, for some number k .
Thus,
k = ( A’B’ / AB ) = ( A’C’ / AC ) = ( B’C’ / BC ) .
Theorem 4.4.1: Similarity of Polygons is an equivalence relation, i.e., it has
the properties of Identity, Reflexivity and Transitivity.
5
Theorem 4.4.2 (The book's Basic Proportionality Theorem):
If a line parallel to one side of a triangle intersects the other two sides in two different
points, then it divides these sides into segments that are proportional.
Given that
B
Line DE || Line AC,
then,
D
E
BD / BA = BE / BC
and
BD / DA = BE / EC.
A
C
(Note: The first equality is proven in the book. The proof given in the textbook
divides the area of the triangle into smaller triangular pieces in two different ways, and
calculating the ratios of the areas these pieces in two ways results in the equality of the
ratios of sides.
The second equality follows from the first algebraically.
It will also be the case that
BD / BE = BA / BC , which also follows algebraically . )
Example: On the triangle to the right, it is given that
BC = 15 and segment DE is parallel to segment AB.
C
8
Determine x and y.
Solution: AC = 11. By Theorem 4.4.2,
D
8 / 11 = x / 15  x = (15)(8) / 11 = 10.91
3
A
x / 8 = y / 3 = 10.91 / 8
x
E
y
B

y = (3)(10.91) / 8 = 4.09
Theorem 4.4.4 is the converse of this Theorem 4.4.2:
If line DE cuts two sides of Δ ABC into proportional segments, then line DE is parallel
to the third side of the triangle.
6
Similarity Conditions of Triangles and The Pythagorean Theorem
Theorem 4.4.5 (The AAA Similarity Condition for Triangles): If, under a correspondence,
the three interior angles of one triangle are congruent to the corresponding angles of a
second triangle, then the triangles are similar, and so corresponding sides are
proportional.
Theorem 4.4.6 (SAS Similarity Condition for Triangles): If, under a correspondence, an
angle of one triangle is congruent to the corresponding angle of another triangle, and if
the corresponding sides that surround these angles are proportional, then the triangles are
similar.
Theorem 4.4.7 (SSS Similarity Condition for Triangles): If, under a correspondence, the
lengths of the three sides of one triangle are proportional to the lengths of the
corresponding sides of another triangle, then the triangles are similar.
Theorem 4.4.8 (The Pythagorean Theorem): If a and b are the lengths of the legs of a
right triangle and if c is the length of the hypotenuse, then
a2 + b2 = c2 .
7
Applications of the Basic Proportionality Theorem:
Angle Bisectors and Ratios of Segment Lengths
Theorem (NIB) 4.1, The "Basic Proportionality Converse" Theorem, Theorem 4.4.2:
(This is actually Theorem 4.4.4)
Given ABC , let D be on AB , and E on AC such that
AD AE

DB EC
or
AD AE

AB AC
then DE is parallel to BC .
Proof. Let F be the point where the line through D parallel to BC
intersects AC .
First, assume
AD AE
.

AB AC
A
Thus, DF is parallel to BC .
Thus, AE  ( AC )
AD
.
AB
D
E
F
By the "Basic Proportionality" Theorem,
AD AF
.

AB AC
AD
Thus, AF  ( AC )
, and so, AE = AF by
AB
C
B
substitution.
 E = F by SMSG Postulates 3 and 4.
Now, assume
Thus DE = DF , which is parallel to BC .
AD AE
. As before, by the "Basic Proportionality" Theorem,

DB EC
AD AF
AE
AF
.



DB FC
EC
FC
Since AC = AE + EC and AC = AF + FC , EC = AC - AE and FC = AC - AF .
Thus, by substitution,
AE
AF
, and so, (AE) (AC – AF) = (AF) (AC – AE) .

AC  AE
AC  AF
 (AE) (AC) – (AE) (AF) = (AF) (AC) – (AF) (AE) . Subtracting (AF) (AE) from both sides,
we conclude that (AE) (AC) = (AF) (AC) . Dividing by (AC),  AE = AF, and as before,
E = F.
Thus DE = DF ,
DE is parallel to BC .
QED
8
Theorem (NIB) 4.2, The "Angle Bisector Third Side Cut" Theorem.
The bisector of the internal angle of a triangle divides the opposite side in the ratio of the
adjacent sides. That is, for a triangle ABC , if D is the point on AC
such that BD is the bisector of ABC in ABC , then
Proof. Suppose BD bisects ABC in ABC .
=
.
At C, construct a line
parallel to BD , intersecting AB at E. Thus,
E
.
||
4
Let ∠ 1 , ∠ 2 , ∠ 3 , and ∠ 4 be the angles as indicated in the
figure.
Since BD is an angle bisector, ∠ 1  ∠ 2 .
Since ∠ 1 and ∠ 4 are corresponding angles formed when
intercepts parallel lines
and
, ∠1  ∠4.
 By transitivity, ∠ 2  ∠ 4 .
B
1 2
Since ∠ 2 and ∠ 3 are alternate interior angles formed when
intercepts parallel lines
and
, ∠ 2  ∠ 3 (AIA Converse).
 By transitivity, ∠ 3  ∠ 4 .
By the "Congruent Base Angles" Theorem, BE = BC .
By the "Basic Proportionality Theorem" applied to  AEC ,
But, BE = BC,
so
=
.
3
A
D
=
C
.
QED
Theorem (NIB) 4.3, The "Angle Bisector Third Side Cut Converse" Theorem:
The converse of the "Angle Bisector Third Side Cut Converse" Theorem is true;
that is, if D is the point on AC in ABC such that
AB
AD
,

BC
DC
then BD is the angle bisector of ABC .
Proof: The proof is left as an exercise.
Example: Given that AC = 25 and that BD is the bisector of ABC in the ABC shown in the
figure, determine the values of x and y.
B
Solution:
By the "Angle Bisector Third Side Cut" Theorem,
Theorem (NIB) 4.2,
x
14
2


y
21
3
Since x  y  25 ,
and so, x 
2
y  y
3

2
y .
3
5
y  25 ,
3
and so y = 15 and x = 10.
21
14
A
x
D
y
C