Transcript Slide 1

Chapter 1
Congruent Triangles
DEFINITION.
Two triangles ABC and  DEF
are said to be congruent if the
following six conditions all hold :
1.A  D
4. AB  DE
2.B  E
5.BC  EF
3.C  F
6. AC  DF
In this case, we write ABC  DEF and we say
that the various congruent angles and segments
"correspond" to each other.
SAS THEOREM (POSTULATE)
If A  D, AB  DE and
then ABC  DEF .
AC  DF ,
If A  D, B  E and AB  DE ,
then ABC  DEF.
Proof of ASA
Assume that in ABC and DEF
A  D, B  E and AB  DE. Consider
now AC and DF. If we knew that they
were congruent, then we would be done:
The two triangles would be congruent by
SAS.
So we assume that they are not equal
and AC is longer. Since AC is longer
there is an interior point G on AC such
that AG  DF . But now ABG and DEF
satisfy: A  D, AB  DE
and AG  DF
So, by SAS, ABG  DEF .
Hence , ABG  E But , by hypothesis
ABC  E
Comparing these two equations, we see that
ABG  ABC
But this is impossible: ABG has to be smaller
than ABC because G is an interior point of
ABC . This is a contradiction.
Hence, our assumption that AC
must
is not congruent to DF
False.
be
Therefore, ABC  DEF
and the proof is concluded.
Isosceles Triangle Theorem
Let ABC be any triangle
1. B  C then AB  AC
2. If
If AB  AC then B  C
Proof. (1):
In this case we have these three
Congruences , B  C , C  B
and BC  CB
Hence, ABC  ACB by ASA (!). So AB AC
as corresponding part of congruent triangles.
Proof. (2):
Now use the congruences,
AB AC ,
AC  AB ,
And A  A .
To conclude that ABC  ACB by SAS.
So B  C .
SSS THEOREM
If AB  DE, BC  EF and AC  DF ,
then ABC  DEF .
Assume in ABC and DEF that
AB  DE, BC  EF and AC. DF
We now construct a point G such that
GBC  E and GCB  F
By ASA, GBC  DEF .
To prove the theorem we will prove that
GBC  ABC
Construct GA .
Since AB DE by assumption,and DE  GB
as corresponding parts of congruent triangles,
we conclude that AB GB , by transitivity.
Hence BAG is an isosceles triangle.
So, BAG  BGA
.
By a similar line of reasoning CAG is an
isosceles triangle and CAG  CGA
Now consider the following relations:
BAG  CAG
CAG  CGA
CGA  BGA
BGA  BAG
Together, these imply that BAG  BAG
This is impossible.Hence, G must coincide with A,
GBC  B  E
and ABC  DEF by SAS.
PROBLEM. To bisect an angle.
Solution. First, use your compass to construct
B and C such that AB  AC .
Then draw an arc with center B, and arc
with center C, both with the same radius .
Let D be their point of intersection.
ThenAD is the angle bisector.
Since
AD  AD,
AB  AC,
and BD  CD,
We may conclude that ABD  ACD by SSS.
Hence, BAD  CAD .
PROBLEM
Given angle A and ray BC , construct an
angle DBC that will be congruent to A .
,
Proof
Since
AP  BC,
AQ  BD,
PQ CD,
We have APQ  BCD by SSS.
Hence
A  DBC .
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 PAQand PBQ .
They must be congruent by SSS. From this
we conclude that APQ  BPQ
Now, let C be the point of intersection of PQ
and AB and consider PAC and PBC
These two triangles have P C as a common
side, they have PA  PB and they have
congruent angles at P, as noted previously.
Hence PAC  PBC. So, PCA  PCB
But, PCA  PCB  180 .
These two equations imply that PCA  90
and PCB  90 ,as claimed .
Perpendicular Bisector of Segment
P
A
B
P
B A
A
Q
B
Q
The Vertical Angel Theorem
Let the lines AB and CD meet at point P, as
shown in the figure. Then APC  BPD
and APD  BPC .
B
C
P
A
D
Proof
Clearly, APC  APD  180 and

Hence,
APD  BPD  180 .

APC  180  APD

BPD  180  APD.

So APC  BPD .The case of APD  BPC
is similar.
THE EXTERIOR ANGLE
THEOREM
In
ABC extend BC to a point D on BC ,
forming the exterior angle ACD . Then
ACD is greater than each of A and B ,
the remote interior angles.
A
F
E
B
C
D
Proof
The first part of the proof will consist of
constructing the picture in figure .
First, let E be the midpoint of AC; then connect
B to E and extend to a point F in the interior of
ACD such that BE  EF. Now consider the
triangles
AEB and CEF .
By construction, they have two of their
respective sides congruent. Moreover,
by the vertical angle theorem AEB  CEF.
Hence AEB  CEF .
This implies that ECF  A . But ECF
is less than ACD . So A  ACD
as claimed . The case of B is similar.
Let ABC be a triangle in which BC is longer
than AC. Then A  B .
or, the greater angle is opposite the greater
side.
A
B
D
C
.
Proof
We may find a point D on BC such that DC  AC .
Since ADC is isosceles,
CAD  CDA
.
By the exterior angle theorem CDA  B,
and, clearly,
A  CAD.
Comparing these three statements, the result
follows.
THEOREM
Let ABC be a triangle in which A  B
Then BC  AC .
or, the greater side is opposite the greater
angle.
A
B
C
then
If the theorem was not true,
either BC  AC or AC  BC .
,then the triangle would be
If BC  AC
A   B. .
and we would get the contradiction isosceles
If AC  BC ,then B  A.
,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
Let ABC be any triangle. We will show that
AB  AC  BC .
Extend the side BA to a point D such that
AD  AC. So BD  AB  AC . now, since
ACD is isosceles, D  DCA
D  DCB . Therefore in BCD
BC is a side opposite a larger angle DCB .
The theorem now follows.