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Transcript
Aim: How to prove triangles are congruent
using a 3rd shortcut: ASA.
Do Now:
Given:
T is the midpoint of PQ, PQ bisects RS, and
RQ  SP. Explain how RTQ  STP.
S
P
T
R
Q
Aim: Triangle Congruence - ASA
Course: Applied Geometry
Do Now
You are given:
T is the midpoint of PQ, PQ bisects RS, and
RQ  SP. Explain how RTQ  STP.
S
P
RQ  SP – we’re told so
T
R
Q
(S  S)
PT  TQ – a midpoint of a
segment cuts the segment into
(S  S)
two congruent parts
RT  TS – a bisector
divides a segment into 2
congruent parts
(S  S)
RTQ  STP because of SSS  SSS
Aim: Triangle Congruence - ASA
Course: Applied Geometry
Sketch 14 – Shortcut #3
A
B
C

Copied 2 angles and included side:
BC  B’C’, B  B’, C  C’
A‘
B’
C’
Measurements
ABC
showed:
Shortcut for
proving
congruence
inCongruence - ASA
Aim: Triangle
triangles:
 A’B’C’
ASA  ASACourse: Applied Geometry
Angle-Side-Angle
III.
ASA = ASA
Two triangles are congruent if two angles
and the included side of one triangle are
equal in measure to two angles and the
included side of the other triangle.
A’
A
B
C
B’
C’
If A =  A', AB = A'B',  B = 
B', then ABC = A'B'C'
If ASA  ASA ,
then the triangles are congruent
Aim: Triangle Congruence - ASA
Course: Applied Geometry
Model Problems
Is the given information sufficient to prove
congruent triangles?
C
YES
C
B
A
D
YES
E
A
NO
Aim: Triangle Congruence - ASA
D
Course: Applied Geometry
B
Model Problems
Name the pair of corresponding sides that
would have to be proved congruent in order
to prove that the triangles are congruent by
ASA.
DCA  CAB
C
D
B
D
DFA 
BFC
F
A
B
D
A
C
C
DB  DB
A
Aim: Triangle Congruence - ASA
B
Course: Applied Geometry
Model Problem
CD and AB are straight
lines which intersect at E.
BA bisects CD. AC  CD,
BD  CD.
Explain how ACE  BDE
using ASA
B
C
D
E
A
C  D –  lines form right angles and all
right angles are  & equal 90o
(A  A)
CE  ED – bisector cuts segment into 2  parts (S  S)
CEA  BED – intersecting straight lines
(A  A)
form vertical angles which are opposite and 
ACE  BDE because of ASA  ASA
Aim: Triangle Congruence - ASA
Course: Applied Geometry
Model Problem
1  2, D is midpoint of EC,
3  4.
Explain how AED  BCD
using ASA
A
1
E
B
3
4
2
C
D
1  2 – Given: we’re told so
(A  A)
ED  DC – a midpoint of a segment cuts the
segment into two congruent parts
(S  S)
3  4 – Given: we’re told so
(A  A)
AED  BCD because of ASA  ASA
Aim: Triangle Congruence - ASA
Course: Applied Geometry
Model Problem
DA is a straight line,
E  B, ED  AB,
FD  DE, CA  AB
Explain how DEF 
ABC using ASA
E
D
C
F
A
B
E  B – Given: we’re told so
(A  A)
ED  AB – Given: we’re told so
(S  S)
EDF  BAC -  lines form right angles and
all right angles are  & equal 90o
(A  A)
DEF Aim:
 ABC
because of ASA

ASA
Triangle Congruence - ASA
Course: Applied Geometry
Aim: Triangle Congruence - ASA
Course: Applied Geometry