Chapter 1 Congruent Triangles. In this case, we write and we say that the various congruent angles and segments "correspond" to each other. DEFINITION.

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 .