MATH 4510/5510: Brown
Congruence-1
Notes on Congruence1
Axiom 1 (C-1). If A and B are distinct points and if A′ is any point, then for each ray r emanating from
A′ there is a unique point B ′ on r such that AB ∼
= A′ B ′ .
Axiom 2 (C-2). If AB ∼
= EF . Moreover, every segment is congruent to
= EF , then CD ∼
= CD and AB ∼
itself.
Axiom 3 (C-3: Segment Addition). If A∗B ∗C, A′ ∗B ′ ∗C ′ , AB ∼
= A′ B ′ , and BC ∼
= B ′ C ′ , then AC ∼
= A′ C ′ .
−−−→
Axiom 4 (C-4). Given any ∠BAC, and given and ray A′ B ′ emanating from a point A′ , then there is a
−−→
←−→
unique ray A′ C ′ on a given side of line A′ B ′ such that ∠B ′ A′ C ′ ∼
= ∠BAC.
Axiom 5 (C-5). If ∠A ∼
= ∠C. Moreover, every angle is congruent to itself.
= ∠C, then ∠B ∼
= ∠B and ∠A ∼
Axiom 6 (C-6: SAS). If two sides and the included angle of one triangle are congruent to two sides and
the included angle of another triangle, then the two triangles are congruent.
Corollary 1 (Corollary to SAS). Given △ABC and segment DE ∼
= AB, there is a unique point F on a
←→
given side of DE such that △ABC ∼
= △DEF .
−−→
←→
Proof. By C-4 there is a unique ray DG on the given side of DE such that ∠BAC ∼
= ∠EDG. By C-1, there
−−→
is a unique point F on DG such that AC ∼
= DF . By SAS, △ABC ∼
= △DEF .
Proposition 10. If in △ABC we have AB ∼
= AC, then ∠B ∼
= ∠C.
Proof. Consider the correspondence of vertices A ↔ A, B ↔ C, C ↔ B. By hypothesis AB ∼
= AC, and by
C-5 ∠A ∼
= ∠C.
= △ACB. By the definition of congruent triangles ∠B ∼
= ∠A. Then by SAS △ABC ∼
Proposition 11 (Segment Subtraction). If A∗B ∗C, D ∗E ∗F , AB ∼
= DE, and AC ∼
= DF , then BC ∼
= EF .
−−→
Proof. Suppose BC 6∼
= EG. By the RAA
= EF (RAA). By C-1, there is a point G on EF such that BC ∼
∼
∼
hypothesis F 6= G. Since AB ∼
DE
by
hypothesis,
and
BC
EG
we
have
AC
=
=
= DG by C-3. By
∼
hypothesis, AC ∼
DF
,
so
by
C-2,
DG
DF
.
By
the
uniqueness
in
C-1
G
=
F
.
This
is
a contradiction, so
=
=
BC ∼
EF
.
=
Proposition 12. Given AC ∼
= DF , then for any point B between A and C, there is a unique point E
between D and F such that AB ∼
= DE.
−−→
Proof. By C-1 there is a unique point E on DF such that AB ∼
= DE. Suppose E is not between D and F
(RAA). By the definition of ray, either E = F or D ∗ F ∗ E. Suppose E = F . B and C are distinct points
−−→
on AB by B − 1, but DE ∼
= AB and DE ∼
= AC. This contradicts C-1, so E 6= F . Now suppose D ∗ F ∗ E.
−→
By C-1 there is a unique point G on the ray opposite to CA such that CG ∼
= F E. By C-3, AG ∼
= DE. This
contradicts the uniqueness part of C-1, since AB ∼
= DE. Hence E is between D and F .
Definition. AB < CD (or CD > AB) means that there exists a point E between C and D such that
AB ∼
= CE.
Proposition 13 (Segment Ordering).
1. Exactly one of the following conditions holds (trichotomy): AB < CD, AB ∼
= CD, or AB > CD.
2. If AB < CD and CD ∼
= EF , then AB < EF .
3. If AB > CD and CD ∼
= EF , then AB > EF .
4. If AB < CD and CD < EF , then AB < EF (transitivity).
Proof.
1 The statements of the propositions and many proofs are taken from the book Euclidean and Non-Euclidean Geometries by
M. Greenberg.
MATH 4510/5510: Brown
Congruence-2
−−→
1. Suppose AB and CD are not congruent. By C-1, there exists a point E on CD such that AB ∼
= CE.
By definition of ray, either C ∗ D ∗ E or C ∗ E ∗ D. If C ∗ E ∗ D, then AB < CD by definition of <.
−−→
Suppose C ∗ D ∗ E. By Proposition 3.12, there is a unique point F on AB between A and B such that
CD ∼
= AF . Then AB > CD by definition of <.
2. Since AB < CD, there is a point P between C and D such that AB ∼
= CP . Since CD ∼
= EF , by
Proposition 3.12, there is a unique point Q between E and F such that CP ∼
= EQ. By C-2, AB ∼
= EQ,
so AB < EF .
3. Since AB > CD, there is a point P between A and B such that AP ∼
= EF . Then
= CD. Suppose CD ∼
by C-2, AP ∼
= EF , so AB > EF .
∼ CP . Since CD < EF , there is
4. Since AB < CD, there is a point P between C and D such that AB =
a point Q between E and F such that CD ∼
= EQ. By Proposition 3, there is a point R between E and
Q such that CP ∼
= ER. By C-2, AB ∼
= ER. Since E ∗ R ∗ Q and E ∗ Q ∗ F , we know by Proposition
3.3 that E ∗ R ∗ F . Hence, AB < EF .
Proposition 14. Supplements of congruent angles are congruent.
−−→
−−→
−−→
Proof. Suppose ∠ABC ∼
= ∠DEF . Let BP be the ray opposite to BA and let EQ be the ray opposite to
−−→
∼ ∠F EQ.
ED. We want to show ∠CBP =
C
P
F
B
A
E
Q
D
Since the points A, C, and P are given arbitrarily on the sides of ∠ABC and ∠CBP , by C-1 we can
choose the points D, F , and Q on ∠DEF and ∠F EQ such that AB ∼
= DE, CB ∼
= F E, and BP ∼
= EQ.
∼
∼
Then △ABC = △DEF by SAS. By the definition of congruent triangles, AC = DF , and ∠A ∼
= ∠D. By
C-3, AP ∼
= DQ. Then again by SAS, △ACP ∼
= △DF Q. By the definition of congruent triangles CP ∼
= FQ
∼
∼
and ∠P = ∠Q. And again by SAS, △CP B = △F QE, so ∠CBP ∼
= ∠F EQ.
Proposition 15.
1. Vertical angles are congruent to each other.
2. An angle congruent to a right angle is a right angle.
Proof.
−−→
−−→
1. By definition two angles are vertical if they allow labeling ∠ABC and ∠DBE where BA and BD are
−−→
−−→
opposite, and BC and BE are opposite.
A
E
B
D
C
MATH 4510/5510: Brown
Congruence-3
∠ABC is the supplement to ∠ABE and ∠DBE is the supplement to ∠ABE. By C-5 ∠ABE ∼
= ∠ABE,
so by Proposition 5 ∠ABC ∼
= ∠DBE.
−→
∼ ∠ABC. Suppose P is a point on the ray opposite to −
2. Suppose ∠ABC is a right angle and ∠DEF =
BC
−−→
and Q is a point on the ray opposite EF . We need to show ∠DEF ∼
= ∠DEQ. Since ∠ABC ∼
= ∠DEF ,
∼
by Proposition 5, their supplements are congruent, i.e. ∠ABP = ∠DEQ. By the definition of right
angle ∠ABC ∼
= ∠DEQ.
= ∠DEQ. Again by C-5 ∠DEF ∼
= ∠ABP , so by C-5 ∠ABC ∼
Proposition 16. For every line l and every point P there exists a line through P perpendicular to l.
Proof. Either P lies on l or it does not. Assume first that P does not lie on l, and let A and B be any
−−→
−−→
two points on l by I-2. By C-4 there is a ray AX such that AX is on the opposite side of l as P and
−−→
∠XAB ∼
= ∠P AB. By C-1 there is a point P ′ on AX such that AP ∼
= AP ′ . Since P and P ′ are on opposite
′
′
sides of l, P P intersects l at a point Q between P and P . If Q = A, ∠P AB and ∠P ′ AB are supplementary.
Since these angles are congruent, they are right angles, so P P ′ ⊥ l. If Q 6= A, then AQ ∼
= AQ by C-1, so
△P AQ ∼
= ∠P ′ QA. Hence P P ′ ⊥ l.
= △P ′ AQ by SAS. By the definition of congruent triangles ∠P QA ∼
Now suppose P lies on l. By Proposition 2.3 there is a point not on l. By the above argument we can
construct a line perpendicular to l through this point, thereby obtaining a right angle. By C-4, there is a
unique ray on a particular side of l emanating from P such that ∠P with one side contained in l is congruent
to a right angle. By Proposition 3.15, ∠P is a right angle. The side of this angle not contained in l is
contained in a line perpendicular to l through P .
Proposition 17 (ASA). Given △ABC and △DEF with ∠A ∼
= ∠D, ∠C ∼
= ∠F , and AC ∼
= DF . Then
∼
△ABC = △DEF .
Proof. By C-1, there is a unique point X on DE such that AB ∼
= DX. By hypothesis ∠A ∼
= ∠D and
∼
∼
AC = DF , so △ABC = △DXF by SAS. By the definition of congruent triangles ∠C ∼
= ∠DF X. By
−−→
−−→
hypothesis ∠C ∼
= ∠DF E, so by the uniqueness of C-4 F E = F X. By Proposition 2.3, E = X. Hence
△ABC ∼
= △DEF .
Proposition 18. If in △ABC we have ∠B ∼
= ∠C, then AB ∼
= AC and △ABC is isosceles.
Proof. Consider the correspondence of vertices A ↔ A, B ↔ C, and C ↔ B. By C-1 BC ∼
= BC. By
∼
hypothesis ∠C ∼
∠B,
so
by
Proposition
3.17,
△ABC
△ACB.
By
the
definition
of
congruent
triangles
=
=
AB ∼
= AC, so by the definition of isosceles triangle △ABC is isosceles.
−−→
−−→
−−→ −−→
−−→
−−→
Proposition 19 (Angle Addition). Given BG between BA and BC, EH between ED and EF , ∠CBG ∼
=
∠F EH, and ∠GBA ∼
= ∠HED. Then ∠ABC ∼
= ∠DEF .
Proof. By the crossbar theorem we may choose G so that A ∗ G ∗ C. By C-1 we may choose D, F , and H so
that AB ∼
= △DEH.
= ∠DEH, so by SAS △ABG ∼
= EH. By hypothesis ∠ABG ∼
= EF , and BG ∼
= DE, BC ∼
∼
∼
Similarly, by hypothesis ∠CBG = ∠F EH, so by SAS △CBG = △F EH. By the definition of congruent
triangles AG ∼
= DH and GC ∼
= HF .
We next need to show D, H, and F are collinear. By the definition of congruent triangles ∠AGB ∼
=
∠DHE and ∠CGB ∼
= ∠F HE. Since A, G, and C are collinear ∠AGB and ∠CGB are supplementary. By
Proposition 5 ∠CGB is congruent to the supplement of ∠DHE. Denote this supplement by ∠EHX. By
−−→
−−→ −−→
C-4 HX is unique, so HX = HF . Then ∠DHE is supplementary to ∠F HE, so F , H, and D are collinear.
−−→
−−→
−−→
Since EH is between ED and EF , H is in the interior of ∠DEF , so D ∗ H ∗ F by Proposition 3.7.
Since AG ∼
= △DEF , so by the definition of
= DF . By SAS △ABC ∼
= HF , by C-3 AC ∼
= DH and GC ∼
∼
congruent triangles ∠ABC = ∠DEF .
−−→
−−→
−−→ −−→
−−→
−−→
Proposition 20 (Angle Subtraction). Given BG between BA and BC, EH between ED and EF , ∠CBG ∼
=
∠F EH, and ∠ABC ∼
= ∠DEF . Then ∠GBA ∼
= ∠HED.
Proof. We proceed as in the proof of Proposition 3.11. Suppose ∠GBA ∼
6 ∠HED (RAA). By C-4 there
=
−−→
←→
−−→
is a unique ray EX on the same side of EH such that ∠GBA ∼
= ∠HEX. By the RAA hypothesis EX 6=
−−→
ED. By hypothesis ∠F EH ∼
= ∠CBG, and by RAA hypothesis ∠XEH ∼
= ∠ABG, so by Proposition 10
∼
∼
∠XEF
. By the uniqueness part of
∠DEF
,
so
by
C-5
∠DEF
∠XEF
.
By
hypothesis
∠ABC
∠ABC ∼
=
=
=
−−→ −−→
∼
C-4 EX = ED, but this is a contradiction, so ∠GBA = ∠HED.
MATH 4510/5510: Brown
Congruence-4
−−→
−−→
−−→
−−→
Lemma 1. Given ∠ABC ∼
= ∠DEF , then for any ray BG between BA and BC, there is a unique ray EH
−−→
−−→
between ED and EF such that ∠CBG ∼
= ∠F EH.
A
D
G
B
H
C
E
F
Proof. By the Crossbar Theorem we can choose G so that A ∗ G ∗ C. By C-1 we can choose points D and F
such that AB ∼
= DE and BC ∼
= EF . By SAS △ABC ∼
= △DEF , and by the definition of congruent triangles
∠BCA ∼
= ∠EF D and AC ∼
= DF . Then by Proposition 3.12 there is a unique point H on DF such that
CG ∼
= F H. Again by SAS, △CBG ∼
= △F EH, so by the definition of congruent triangles ∠CBG ∼
= ∠F EH.
−−→
−−→
−−→
We only need to show that EH is between ED and EF . Since H is on DF , H is between D and F , so by
−−→
−−→
−−→
Proposition 3.7 EH is between ED and EF .
−−→
−−→
−−→
∼ ∠GEF .
Definition. ∠ABC < ∠DEF means there exists a ray EG between ED and EF such that ∠ABC =
Proposition 21 (Ordering of Angles).
1. Exactly one of the following conditions holds (trichotomy): ∠P < ∠Q, ∠P ∼
= ∠Q, or ∠P > ∠Q.
2. If ∠P < ∠Q and ∠Q ∼
= ∠R, then ∠P < ∠R.
3. If ∠P > ∠Q and ∠Q ∼
= ∠R, then ∠P > ∠R.
4. If ∠P < ∠Q and ∠Q < ∠R, then ∠P < ∠R. (transitivity).
Proof. This proof is very similar to the proof of Proposition 3.13. For labeling purposes we say ∠P = ∠ABC,
∠Q = ∠DEF , and ∠R = ∠GHI.
−−→
←→
1. Suppose ∠ABC ∼
6 ∠DEF . By C-4 there exists a unique ray EX on the same side of EF as D such
=
−−→
−−→
−−→
−−→
−−→
−−→
−−→
that ∠ABC ∼
. EX either is between EF and ED or EX is not between EF and ED. If EX
= ∠XEF
−−→
−−→
is between EF and ED, then ∠ABC < ∠DEF .
−−→
←→
−−→
−−→
−−→
Suppose EX is not between EF and ED. Since X is on the same side of EF as D and EX is not
−−→
−−→
←→
between EF and ED, we know that X and F are on opposite sides of ED. By Lemma 2 and Corollary
−−→
←→
−−→
1 every point except D on DF is on the opposite side of ED as every point of EX, so segment DF
−−→
−−→
−−→
does not intersect EX. By a similar argument with the ray opposite EX and F D, we can show that
←→
←→
segment DF does not meet EX. Hence D is on the same side of EX as F , so D is interior to ∠XEF .
−−→
−−→
−−→
−−→
−−→
−−→
Then ED is between EX and EF . By Lemma 4 there exists a unique ray BY between BA and BC
such that ∠DEF ∼
= ∠Y BC. Hence ∠ABC > ∠DEF .
−−→
−−→
−−→
2. Since ∠ABC < ∠DEF , there exists a ray EX between ED and EF such that ∠ABC ∼
= ∠XEF .
−−→
−−→
−→
∼
By Lemma 4 there is a unique ray HY between HG and HI such that ∠Y HI = ∠XEF . By C-5
∠ABC ∼
= ∠Y HI, so ∠ABC < ∠GHI.
−−→
−−→
−−→
3. Since ∠ABC > ∠DEF , there exists a ray BX between BA and BC such that ∠XBC ∼
= ∠DEF . By
C-5 ∠XBC ∼
= ∠GHI, so ∠ABC > ∠GHI.
−−→
4. Suppose ∠ABC < ∠DEF and ∠DEF < ∠GHI. Since ∠ABC < ∠DEF , there is a unique ray EX
−
−→
−−→
−−→
between ED and EF such that ∠ABC ∼
= ∠XEF . Since ∠DEF < ∠GHI, there is a unique ray HY
−−→
−→
−
−
→
−
−
→
between HG and HI such that ∠DEF ∼
= ∠Y HI. By Lemma 4, there is a unique ray HZ between HY
−→
−→ −−→ −−→
−→ −−→ −−→
and HI such that ∠XEF ∼
= ∠ZHI. By C-5 ∠ABC ∼
= ∠ZHI. Since HI ∗ HZ ∗ HY and HI ∗ HY ∗ HZ,
−→ −−→ −−→
by Lemma 3 HI ∗ HZ ∗ HZ. Then by definition ∠ABC < ∠GHI.
MATH 4510/5510: Brown
Congruence-5
Proposition 22 (SSS). Given △ABC and △DEF . If AB ∼
= DE, BC ∼
= EF , and AC ∼
= DF , then
△ABC ∼
= △DEF .
←→
Proof. By Corollary 4, since AC ∼
= DF we can pick a point G uniquely on the opposite side of AC as B
such that △DEF ∼
= GC. Then by
= AG and EF ∼
= △AGC. By the definition of congruent triangles DE ∼
∼
∼
∼
C-2 AB = AG and BC = GC. We will show that △ABC = △AGC. Since B and G are on opposite sides of
←→
←→
AC, segment BG intersects AC at X. By B-3, X ∗ A ∗ C, A = X, or A ∗ X ∗ C. The circumstances X = C
and A ∗ C ∗ X are equivalent to X = A and X ∗ A ∗ C, respectively.
Suppose X ∗ A ∗ C. Consider △CBG. Since BC ∼
= CG, △CBG, by Proposition 3.10 ∠CBG ∼
= ∠CGB.
∼
Now consider △ABG. Since AB = AG, by Proposition 3.10 ∠ABG ∼
= ∠AGB. Then by Proposition 3.20
(angle subtraction) ∠CBA ∼
= ∠CGA. By SAS △ABC ∼
= △AGC.
Suppose A = X. Since CG ∼
= CB in △CBG, by Proposition 10 ∠B ∼
= ∠G. Then by SAS △ABC ∼
=
△AGC.
Suppose A ∗ X ∗ C. Consider △CBG. Since BC ∼
= CG, △CBG, by Proposition 3.10 ∠CBG ∼
= ∠CGB.
∼
Now consider △ABG. Since AB = AG, by Proposition 3.10 ∠ABG ∼
= ∠AGB. Then by Proposition 3.19
(angle addition) ∠CBA ∼
= ∠CGA. By SAS △ABC ∼
= △AGC.
In all three cases △ABC ∼
= △AGC. By the definition of congruent triangles ∠B ∼
= ∠G. Since △DEF ∼
=
∼
∼
∼
△AGC, ∠E = ∠G. By C-5 ∠B = ∠E, so by SAS △ABC = △DEF .
Proposition 23. All right angles are congruent to each other.
Proof. Suppose ∠DAB ∼
6
= ∠DAC and ∠HEF ∼
= ∠HEG are two pairs of right angles. Assume ∠DAB ∼
=
∠HEF (RAA). By Proposition 3.21 (a), either ∠DAB > ∠HEF or ∠DAB < ∠HEF . Without loss of
−−→
−−→
−−→
generality suppose ∠DAB > ∠HEF . Then there is AX between AB and AD such that ∠XAB ∼
= ∠HEF .
By Proposition 3.14 ∠XAC ∼
= ∠HEG.
Since ∠DAC ∼
= ∠DAB by hypothesis, and ∠DAB > ∠HEF by by RAA hypothesis, we have ∠DAC >
∠HEF by Proposition 3.21 (c). Since ∠HEF ∼
= ∠HEG by hypothesis, we have again by Proposition 3.21
(c) ∠DAC > ∠HEG. From above since ∠XAC ∼
= ∠HEG we have ∠DAC > ∠XAC by Proposition 3.21
−−→
−−→
−−→
−−→
−−→
−→
(c). By Proposition 3.8 (c), since AX is between AB and AD, we know that AD is between AX and AC,
−→
−−→
since AC is the ray opposite to AB. Then ∠DAC < ∠XAB, but this contradicts Proposition 3.21 (a).
Hence ∠DAB ∼
= ∠HEF .