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October 18, 2015
Geometry.
Similarity and homothety. Theorems and problems.
Definition. Two figures are homothetic
with respect to a point , if for each point
of one figure there is a corresponding
point belonging to the other figure,
such that lies on the line
at a
distance
from
point , and vice versa, for each point
of the second figure there is a
corresponding point belonging to the
first figure, such that lies on the line
at a distance
Homothety
D’
O
from point .
D
E
F
A
C’
C
E’
B
F’
B’
A’
O
C
C’
Here the positive number is called the
homothety (or similarity) coefficient. Homothetic
figures are similar. The transformation of one
A’
figure (e.g. multilateral ABCDEF) into the figure
A B C D E F is called homothety, or similarity transformation.
B
A
B’
Thales Theorem Corollary 1. The corresponding segments (e.g. sides) of the
homothetic figures are parallel.
Thales Theorem Corollary 2. The ratio of the corresponding elements (e.g.
sides) of the homothetic figures equals .
Exercise. What is the ratio of the areas of two similar (homothetic) figures?
Definition. Consider triangles, or polygons, such that angles of one of them are
congruent to the respective angles of the other(s). Sides which are adjacent to
the congruent angles are called homologous. In triangles, sides opposite to the
congruent angles are also homologous.
Generalized Pythagorean Theorem 2.
C
Theorem 2. For three homologous segments,
,
and
belonging to the similar
right triangles
,
and
, where
is the altitude of the triangle
drawn
to its hypotenuse
, the following holds,
lACD
lABC
A
lCBD
D
B
Proof. If we square the similarity relation for the homologous segments,
, where
hypotenuse of the triangle
,
and
are the legs and the
, we obtain,
. Using the
property of a proportion, we may then write,
Pythagorean theorem for the right triangle
immediately obtain
.
, wherefrom, by
,
, we
Theorem 1. If three similar polygons, P, Q and R with
areas , and are constructed on legs , and
hypotenuse , respectively, of a right triangle, then,
Proof. The areas of similar polygons on the sides of a
right triangle satisfy
and
Q
P
ba
a c
, or,
R
. Using the property of a proportion, we
may then write,
, wherefrom, using the
Pythagorean theorem for the right triangle
immediately obtain
.
,
, we
Property of the bisector.
Theorem (property of the bisector). The bisector of any angle of a triangle
divides the opposite side into parts proportional to the
adjacent sides,
D
,
,
Proof. Consider the bisector BB . Draw line parallel to
BB from the vertex C, which intercepts the extension of
the side AB at a point D. Angles B BC and BCD have
parallel sides and therefore are congruent. Similarly are
congruent ABB and CDB. Hence, triangle CBD is
isosceles, and |BD| = |BC|. Now, applying the intercept
theorem to the triangles ABB and ACD, we obtain
B
A’
C’
O
B’
A
C
.
Theorem (property of the external
D
bisector). The bisector of the
exterior angle of a triangle
intercepts the opposite side at a
point (D in the Figure) such that
the distances from this point to the
vertices of the triangle belonging
to the same line are proportional
to the lateral sides of the triangle.
B
C’
Proof. Draw line parallel to AD
from the vertex B, which intercepts
A
the side AC at a point B . Angles ABB and DAB have parallel sides and
therefore are congruent. Similarly, we see that angles AC B and ABB are
congruent, and, therefore, AB
AB . Applying the intercept theorem, we
obtain,
.
C
B’
Selected problems on similar triangles (from last homework).
Problem 1 (5). In the isosceles triangle
point
divides the side
into segments such that
. If CH is the altitude of the triangle and point O is the
intersection of
and
, find the ratio
to
.
Solution. First, let us perform a supplementary
construction by drawing the segment
parallel to
,
, where point belongs to the side
, and point
to
and the altitude
. Notice the similar triangles,
, which implies,
theorem,
2
D
F
E
O
1
. By Thales
, and
that
C
, so
.
A
H
B
, because
. Therefore, the sought ratio is,
.
Problem 2 (6). Point D belongs to the continuation of side CB of the triangle
ABC such that |BD| = |BC|. Point F belongs to side AC, and |FC| = 3|AF|.
Segment DF intercepts side AB at point O. Find the ratio |AO|:|OB|.
Solution. First, let us perform a supplementary construction by drawing the
segment
parallel to ,
, where belongs to the side
of the
triangle
.
is the mid-line of the
D
triangle
, and, by Thales, also of
and
. Therefore,
and
,
, so
E
G
B
. On the other hand, again,
by Thales, or, noting similar triangles
,
.
O
A
F
C
Commensurate and incommensurate segments. The Euclidean algorithm.
Definition. Two segments, and , are
commensurate if there exists a third
s
segment, , such that it is contained in each
b=ms
of the first two segments a whole numbers of
times with no remainder.
s
,
,
a=ns
,
The segment is called a common measure of the segments and . The
concept of commensurability is similar to that of the common divisor for
integers. It can be extended to any two quantities of the same denomination –
two angles, two arcs of the same radius, or two weights.
The greatest common measure.
If a common measure of two segments and is sub-divided into two, three,
or, generally, any number of equal smaller segments, these smaller segments
are also common measures of the segments and . In this way, an infinite set
of common measures, decreasing in length, can be obtained,
,
. Since
any common measure is less than the smaller segment,
, there must be
the largest among the common measures, which is called the greatest
common measure.
Finding the greatest common measure (GCM) is done by the method of
consecutive exhaustion called Euclidean algorithm. It is similar to the method
of consecutive division used for finding the greatest common divisor in
arithmetic. The method is based on the following theorem.
Theorem. Two segments and are commensurate, if and only if the smaller
segment, , is contained in the greater one a whole number of times with no
remainder, or with a remainder,
, which is commensurate with the
smaller segment, .
,
,
.
The greatest common measure of two segments is also the greatest common
measure of the smaller segment and the remainder, or there is no remainder.
Proof. First, consider the necessary condition. Let and be commensurate,
, ,
, and
. Let be their greatest
common measure. Then, either
(
) and segment is contained in
a whole number of times with no remainder, being the GCM of the two
segments, or,
,
. Then,
,
where
, and, therefore,
,
, which shows
that and are commensurate. The sufficiency follows from the observations
that (i) if segment is contained in a whole number of times with no
remainder, then the segments are commensurate, and is the greatest
common measure of the two, while (ii) if
, and and are
commensurate with the greatest common measure ,
,
, then
,
, and and are also
commensurate with the same GCM.
The Euclidean algorithm.
In order to find the GCM of the two segments, and , we can proceed as
follows. First, using a compass exhaust the greater segment, marking on it the
smaller segment as many times as possible, until the remainder is smaller
than the smaller segment, , or there is no remainder. According to
Archimedes exhaustion axiom, these are the only two possible outcomes.
Following the above theorem, the problem now reduces to finding the GCM of
this remainder, , and the smaller segment, . We now repeat the same
procedure, exhausting segment with , and again, there is either no
remainder and is the GCM of and , or there is a remainder
. The
problem is then reduced to finding the GCM of a pair of even smaller
segments, and , and so on. If segments and are commensurate and
their GCM, , exists, then this process will end after some number of steps,
namely, on step where
. Indeed, all remainders in this process are
multiples of ,
,
and
is the
decreasing sequence of natural numbers, which necessarily terminates, since
any non-empty set of positive integers has the smallest number “principle of
the smallest integer” . If the procedure never terminates, then segments
have no common measure and are incommensurate.
and
Example. The hypotenuse of an isosceles right triangle is incommensurate to
its leg. Or, equivalently, the diagonal of a square is incommensurate to its side.
Proof. Consider the isosceles right triangle
B
ABC shown in the Figure. Because the
E
hypotenuse is less than twice the leg by the
triangle inequality, the leg can only fit once in
the hypotenuse, this is marked by the segment
AD. Let the perpendicular to the hypotenuse at
C
D
A
point D intercept leg BC at point E. Triangle
BDE is also isosceles. This is because angles BDE and DBE supplement equal
angles ADB and ABD to 90 degrees, and therefore are also equal. Triangle CDE
is an isosceles right triangle, similar to ABC. Its leg |DC|=|AC|-|AB|=|DE|=|BE|
is a remainder of subtracting the leg |AB|=|AD| from the hypotenuse, |AC|,
while the hypotenuse, |CE| =|BC|-|BE|=|BC|-|DC|, is the remainder of
subtracting this remainder from the leg |AB|=|BC|. Hence, on the second step
of the Euclidean algorithm we arrive at the same problem as the initial one,
only scaled down by some overall factor. Obviously, this process never ends,
and therefore the hypotenuse |AC| and the leg |AB| are incommensurate.