Transcript Math 258 - University of Saskatchewan

Chapter 4 : Similar Triangles
• Informally, similar triangles can be thought of as having
the same shape but different sizes. If you had a picture of a
triangle and you enlarged it, the result would be similar to
the original triangle. Here is the precise definition(Fig.4.1)
•
Definition. Two triangles ABCand DEF are said to be
similar with ration k if the following conditions hold for
some positive real number k:
1. A  D
2. B  E
3. C  F
4. DE = k . AB
5. EF = k . BC
6. DF = k . AC
• The theory of similar triangles is similar in many ways to
the theory of congruent triangles. In particular, we have
analogues of the three basic congruence theorems ASA,
SAS, and SSS. These similarity theorems are the main
subject of this section.
•
Theorem (ASA for similar triangles). If A  D, B  E,
and if DE = k . AB, then ABC and DEF are similar
with ratio k.
• Proof. The proof will be based on comparing the areas of
two triangles. SinceA  D we know from Section 3.3 that
the ratio of the areas is the same as the ratio of the products
of the including sides. Hence
areaDEF DE .DF
DF

 k.
.
areaABC AB. AC
AC
( 4.1)
• Since DE = k . AB Likewise, since B  E we get
areaDEF DE .EF
EF

 k.
.
areaABC AB.BC
BC
( 4 .2 )
•
We also know that C  F , since A  D, B  E , and
the sum of three angles in each triangle is 180 . Hence
areaDEF DF .EF

.
areaABC AC.BC
(4.3)
• We now compare these three equations. Comparing (4.1)
and (4.3), we yields
DF DF .EF
k.

.
AC AC.BC
• We can cancel DF from both sides and then multiply both
AC
sides by BC. The result is EF = k . BC. Likewise, if we
work with (4.2) and (4.3), we obtain DF = k . AC. This
completes the proof.
• The other two basic similarity theorems follow from this
one in combination with the appropriate theorems for
congruent triangles.
• Theorem (SAS for similar triangles). If A  D ,
DE = k . AB and DF = k . AC, then ABC and DEF are
similar with ratio k.
•
Proof. We construct a third triangle DE F such that ,
E   DE
, D   A 
and
. DSince
DE = k . AB, it follows that

E   B
. Thus by ASA
for
and
 similar
DE
k.AB triangles,
are similarwith
Hence
. But DF is also
E F 
ABCratio k.
D
equal to
which implies
. Now compare
DkF. AC,
 k.AC
F 
DEso
and
,
,
, and
DEF
DF  : DF
E 

D triangles
D DE are
these
 Dcongruent
DFby SAS.
DF  Hence
Eby
 ASA
E  for
B
and the theorem now follows
similar triangles.
• Theorem (SSS for similar triangles). If DE = k . AB,
EF = k . BC, and DF = k . AC, then
and
ABC
are similar with ratio k.
DEF
• Proof. Construct a third triangle DE F such that D  A
, DE  DE  and DF  DF. The proof is now parallel to the
previous proof: ABC and DE F  are similar with
ratio k by SAS for similar triangles. This in turn implies
that EF  E F  and consequently DEF  DE F by
SSS.
• In most applications of similar triangles we will use a
weaker property than the one we have been studying. We
say that ABC and DEF are similar-without specifying
k-if for some k the triangles are similar with ratio k. If you
look at the definition at the beginning of this chapter, you
will see that k can be calculated from the triangles
themselves as
k
DE
EF
DF
, or k 
, or k 
.
AB
BC
AC
• This leads the following definition.
• Definition. Two triangles ABC and DEF are said to be
F if the three
similar if A  D, B  E , C  
and
numbers DE , EF ,and DF are all equal; that is,
AB BC
AC
DE EF DF


AB BC AC
• In this case we write ABC ~ DEF.
• We may now rewrite our three basic theorems using this
new definition, without mention of the ratio k.
• Theorem (ASA for similar triangles). (also called the AA
theorem) If A  D and B  E then ABC ~ DEF .
• Theorem (SAS for similar triangles). If A  D and
,DE  DF , then ABC ~ DEF.
AB
AC
• Theorem (SSS for similar triangles). If
then ABC ~ DEF.
DE DF EF


AB AC BC
,
• Consider an example. Let AB = 6, BC = 12, and AC =15
and let DE = 60, EF =120, and DF = 150. Then
by SSS for similar triangles. We see this by comparing the
sides of ABC with the sides of DEF and noting that
each side of DEF is 10 times bigger than the
corresponding side of ABC (k = 10). However, we cold
also compare the sides of ABC to each other and the
sides of DEF to each other and compare the ratios:
• BC is twice as long as AB and EF is twice as long as DE;
1
AC is 2 2 times as long as AB, etc. Algebraically, of
DE DF
course, it is obvious that the equation AB  AC is
DE AB
equivalent to the equation DF  AC , and likewise for the
other ratios. Hence we may rewrite the last two theorems.
• Theorem (SAS for similar triangles). If A  D and
, AB  DE then ABC ~ DEF.
AC
DF
• Theorem (SSS for similar triangles). If
, AB  DE , then ABC ~ DEF .
BC
EF
AB DE

AC DF
and
Applications to Constructions
• In this section we construct geometric solutions to three
types of algebraic equations. The first is x  b , the second
type is the pair of simultaneous equationsa c
x b
x  y  a and
 ,
y c
• And the third type is similar to the second with x – y
replacing x + y.
• Problem. Given line segments of lengths a, b, and c,
construct a line segment of length x such that x  b .
a
c
• Solution. Let  2 be any line. As in Fig. 4.4, construct
points A, B, C on  2such that AB = b and AC = c. Let 1
be another line through A and construct D such that
AD = a. Draw CD and construct a line through B, parallel
to CD , which intersects 1 at the point X. Then AX = x is
the required distance.
• Proof. Since CD BX , BCD  ABX since they are
corresponding angles, and A  A . Hence, by the AA
theorem, ABX ~ ACD. So AX  AB , or x  b .
AD
AC
a
c
• Problem. Given a line segment AB of length a and two
other segments with lengths b and c, find a point C in AB
such that AC  b .
CB
c
• Solution. Construct a line through A (not equal to AB ) and
a line parallel to it through B. Construct points D on the
first line and E on the second, as in Fig. 4.5, such that
AD = b, BE = c, and such that D and E are on opposite
sides of AB . Then C will be the intersection point of DE
and AB . Moreover, if AC = x and CB = y, then x and y
• are solutions to the system of equations x + y = a, x  b .
y
c
• Proof. Since AD BE , D  E as alternate interior angles.
Also , ACD  BCE by the vertical angle theorem.
Hence ACD ~ BCE by AA. So AC  AD  b .
BC
BE
c
• Problem. Given a line segment AB and two other
segments with lengths b and c, with b  c , find a point C
on AB such that AC  b .
BC
c
• Solution. As in the previous problem, we construct points
D and E such that AD BE , AD = b, BE = c, but in this
case we take D and E to be on the same side of AB (Fig.
4.6.).
• Proof. As before, the proof follows from ACD ~ BCE .
• As a further application of these ideas we show how to
multiply a line segment by a rational number. For example,
say we would like to find a line segment AC two-fifths as
long as given segment AB . We draw another line through
A, as in Fig. 4.7, and choose any point A1 on that line.
Then, moving further along the line we construct A2 , A3 , A4 ,
and A5 such that AA1  A1 A2  A2 A3  A3 A4  A4 A5 . Join
points A5 and B and draw a line  through A2 parallel to A5 B
. Then C will be the intersection point of  with AB
2
By our first construction, AC  . AB.
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