Chapter 1
Congruent Triangles
DA.1
FC .3
DEAB .4
EFBC .5
DFAC .6
In this case, we write and we say that the various congruent angles and segments "correspond" to each other.
EB .2
DEFABC
DEFINITION. Two triangles ABC and DEF
are said to be congruent if the following six conditions all
hold:
SAS THEOREM (POSTULATE) If and ,
then .
DEAB,DA DFAC DEFABC
If and ,
then .
EBDA ,DEFABC
DEAB
Proof of ASA
Assume that in and and . Consider
now and . If we knew that they were congruent, then we would be done:
The two triangles would be congruent by SAS.
ABC DEFEB,DA DEAB
AC DF
So we assume that they are not equal and is longer. Since is longer there is an interior point G on such that . But now and
AC AC
DFAG ABG DEF
,DA DEAB satisfy:
DFAG
and
DEFABG So, by SAS, .
AC
EABG EABC
ABCABG
Hence , But , by hypothesis
ABG
ABC
But this is impossible: has to be smaller
ABC
DF
than because G is an interior point of
Hence,
This is a contradiction. ACour assumption that
is not congruent to must be False.
Therefore, DEFABC and the proof is concluded.
.
Comparing these two equations, we see that
Let be any triangle
CB ACAB 1 .If
then
ACAB CB 2 .If then
ABC
Isosceles Triangle Theorem
Proof. (1):
In this case we have these three Congruences , BC , CB
and CBBC
So Hence , ACAB
as corresponding part of congruent triangles.
by ASA.)!( ACBABC
Proof. (2):
Now use the congruences,
,
,
And .
To conclude that by SAS.
So .
ACAB
ABAC AA
ACBABC CB
SSS THEOREM
If and ,
then .
EFBC,DEAB DFAC DEFABC
Assume in andABC
G
DEF
We now construct a point such that
DEAB EFBC DFAC
that
, and
EGBC FGCB and
By ASA, DEFGBC
.
.
To prove the theorem we will prove that
ABCGBC
Construct GA
.
DEAB by assumption, Since and GBDE
as corresponding parts of congruent triangles, GBAB , by transitivity .
Hence BAG is an isosceles triangle.
So, BGABAG
.
we conclude that
CAGCGACAG
By a similar line of reasoningisosceles triangle and
is an
Now consider the following relations: CAGBAG CGACAG BGACGA BAGBGA
Together , these imply that BAGBAG
This is impossible. Hence, G must coincide with A, EBGBC
and DEFABC by SAS.
PROBLEM. To bisect an angle.
Solution. First, use your compass to construct
ACAB B and C such that .
Then draw an arc with center B, and arcwith center C, both with the same radius. Let D be their point of intersection.
ThenAD is the angle bisector.
Since ,ADAD ,ACAB
and ,CDBD We may conclude that ACDABD by SSS.
Hence, CADBAD
.
PROBLEM
Given angle and ray , construct an A BC
,
DBC Aangle that will be congruent to .
Proof
Since ,BCAP ,BDAQ
,CDPQ We have BCDAPQ by SSS.
Hence DBCA .
PROBLEM
Given a line and a point P not on . construct a line that passes through P
and is perpendicular to
.
Proof Consider the two triangles and . PAQ
They must be congruent by SSS. From this
PBQ
we conclude that BPQAPQ Now, let C be the point of intersection of PQ
and AB and consider PAC and PBC
These two triangles have PC as a common
side, they have PBPA and they have
congruent angles at P, as noted previously.
Hence PBCPAC So, PCBPCA But, 180PCBPCA
These two equations imply that 90PCA
and 90PCB ,as claimed .
.
.
A B A B A B
P P
Q Q
Perpendicular Bisector of Segment
The Vertical Angel Theorem
Let the lines and meet at point P, as shown in the figure. Then and
AB CDBPDAPC
.BPCAPD B
C
AD
P
Proof
Clearly, and Hence,
So The case of is similar.
180 APDAPC.180 BPDAPD
.180
180
APDBPD
APDAPC
.BPDAPC BPCAPD
THE EXTERIOR ANGLE THEOREM
In extend to a point D on ABC BC BC ,
forming the exterior angle ACD . Then
ACD A is greater than each of and B ,
the remote interior angles.
B
A
E
F
C D
Proof
The first part of the proof will consist of
constructing the picture in figure .
First, let E be the midpoint of
B to E and
AC ; then connect
extend to a point F in the interior of
ACD such that EFBE Now consider the .
AEB CEFtriangles and .
By construction, they have two of their
respective sides congruent. Moreover,
by the vertical angle theorem CEFAEB .
Hence CEFAEB .
This implies that AECF ECF. But
is less than ACD . So ACDA
as claimed The case of is similar. . B
be a triangle in which ABC BCLet is longer
AC BA than
.
. Then .
or, the greater angle is opposite the greater side.
B CD
A
Proof
We may find a point D on BC such that ACDC .
Since ADC is isosceles , CDACAD .By the exterior angle theorem ,BCDA and, clearly, .CADA
Comparing these three statements, the result follows.
THEOREM
Let be a triangle in which ABC BA
Then ACBC .
or, the greater side is opposite the greater angle.
B C
A
If the theorem was not true, then
ACBC BCAC either or
.
.
If ACBC ,then the triangle would be isosceles
and we would get the contradiction BA .
If ,then BCAC AB .
,which is also a contradiction.
THEOREM
In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. D
A
B C
be any triangle. We will show that
BCACAB .BA
ACAD to a point D such that
. So ACABBD . now, since
ACD DCAD is isosceles,
so DCBD
. Therefore in BCD is a side opposite a larger angle BC
Extend the side
DCB
ABCLet
.
The theorem now follows .