Download Triangle Congruence

I Have CPCTC
by Karadimos, MD
I woke up this morning not feeling like I should.
The doctor told me it's not good.
Something has happened to me.
I came down with CPCTC.
Doc said, "When your triangles became identical,
your corresponding parts measured equal."
Triangle
Congruence
I said, "When corresponding parts were the
same,
congruent triangles were to blame."
The bad news is,
CPCTC is very contagious.
The good news is,
you can use it to be courageous.
Solving proofs can be tough.
SSS, SAS, ASA, AAS isn't always enough.
CPCTC is the next device.
You'll hear Karadimos, MD give that advice.
The test for CPCTC,
is to examine the geometry.
Congruent triangles is the start.
CPCTC is the very next part.
To find the cure for the CPCTC blues,
wait for non-congruent triangles to hit the news.
Doctors have no pills,
for my CPCTC ills.
Triangle Congruence Theorem
In geometry, CPCTC is
the abbreviation of a
theorem involving
congruent triangles.
CPCTC stands for
Corresponding Parts of
Congruent Triangles are
Congruent. CPCTC states
that if two or more
triangles are congruent,
then all of their
corresponding parts are
congruent as well.
If ABC  DEF then,
AB  DE , BC  EF , AC  DF
A  D, B  E , C  F
Notes: Example proofs
Given : AB  DE ;B  E , BC  EF
Prove : CA  FD
Statements
Reasons
1. AB  DE (S)
1. Given
2.B  E (A)
2. Given
3.BC  EF (S)
3. Given
4.ABC  DEF
4. SAS  SAS
5.CA  FD
5. CPCTC
Another Example proof
Given:E is the midpoint of BD
BD bisects AC
Prove : B  D
Statements
Reasons
1. E is the midpoint of BD
1. Given
BD bisects AC
2. BE  DE
(S)
3.BEA  DEC
4. AE  CE
5.BEA  DEC
6.B  D
(A)
(S)
2. A midpoint divides a segment into
2 congruent segments.
3. Vertical angles are congruent.
4.A segment bisector divides a
segment into 2 congruent segments.
5.SAS  SAS
6. CPCTC
Given : AF is the perpendicu lar bisector of GI ; 1  2; FO  FT
Prove : O  T
Statements
G
Reasons
1. AF is the perpendicu lar bisector of GI
1. GIVEN
2. 1  2
2. GIVEN
1
F
I
3.FO  FT (S)
4.GF  IF (S)
5. GFA and IFA are right angles.
3. GIVEN
4. Definition of segment bisector
5. Definition of perpendicular lines
6. GFA  IFA
6. All right angles are congruent.
7. mGFA  m1  mGFO
mIFA  m2  mIFT
7. Partition postulate
8. m1  mGFO  m2  mIFT 8. Substitution Property
9. GFO  IFT (A)
9. Subtraction Property of equality
10. GFO  IFT
10. SAS  SAS
11. O  T
11. CPCTC
O
A
2
T
Why was the obtuse angle upset?
Because he was never right!!!!
What do you get when you cross
geometry with McDonalds?
A plane cheeseburger!