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SECTION 4-4
Proving Triangles Congruent: ASA, AAS
ESSENTIAL QUESTIONS
How do you use the ASA Postulate to test for triangle
congruence?
How do you use the AAS Postulate to test for triangle
congruence?
VOCABULARY
1. Included Side:
Postulate 4.3 - Angle-Side-Angle (ASA) Congruence:
Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
VOCABULARY
1. Included Side: The side between two consecutive
angles in a triangle
Postulate 4.3 - Angle-Side-Angle (ASA) Congruence:
Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
VOCABULARY
1. Included Side: The side between two consecutive
angles in a triangle
Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: If
two angles and the included side of one triangle are
congruent to two angles and included side of a
second triangle, then the triangles are congruent
Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
VOCABULARY
1. Included Side: The side between two consecutive
angles in a triangle
Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: If
two angles and the included side of one triangle are
congruent to two angles and included side of a
second triangle, then the triangles are congruent
Theorem 4.5 - Angle-Angle-Side (AAS) Congruence: If
two angles and the nonincluded side of one triangle
are congruent to the corresponding angles and
nonincluded side of a second triangle, then the
triangles are congruent
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
Given: L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
Given: L is the midpoint of WE,WR ! ED
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm.
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm.
4. ∠LWR ≅ ∠LED
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm.
4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm.
4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm
5. △WRL ≅△EDL
1. L is the midpoint of WE,WR ! ED
EXAMPLE 1
Prove the following.
Prove: △WRL ≅△EDL
1. Given
Given: L is the midpoint of WE,WR ! ED
2. WL ≅ EL 2. Midpoint Thm.
3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm.
4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm
5. △WRL ≅△EDL 5. ASA
1. L is the midpoint of WE,WR ! ED
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N 2. Reflexive
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N 2. Reflexive
3. △JNM ≅△KNL
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N 2. Reflexive
3. △JNM ≅△KNL 3. AAS
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N 2. Reflexive
3. △JNM ≅△KNL 3. AAS
4. LN ≅ MN
EXAMPLE 2
Prove the following.
Prove: LN ≅ MN
Given: ∠NKL ≅ ∠NJM, KL ≅ JM
1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
2. ∠N ≅ ∠N 2. Reflexive
3. △JNM ≅△KNL 3. AAS
4. LN ≅ MN 4. Corresponding Parts
of Congruent Triangles
are Congruent (CPCTC)
EXAMPLE 3
On a template design for a certain envelope, the top
and bottom flaps are isosceles triangles with congruent
bases and base angles. If EV = 8 cm and the height of
the isosceles triangle is 3 cm, find PO.
EXAMPLE 3
On a template design for a certain envelope, the top
and bottom flaps are isosceles triangles with congruent
bases and base angles. If EV = 8 cm and the height of
the isosceles triangle is 3 cm, find PO.
EV ≅ PL, so each segment has a measure
of 8 cm. If an auxiliary line is drawn
from point O perpendicular to PL, you
will have a right triangle formed.
EXAMPLE 3
On a template design for a certain envelope, the top
and bottom flaps are isosceles triangles with congruent
bases and base angles. If EV = 8 cm and the height of
the isosceles triangle is 3 cm, find PO.
EV ≅ PL, so each segment has a measure
of 8 cm. If an auxiliary line is drawn
from point O perpendicular to PL, you
will have a right triangle formed.
In the right triangle, we have one leg (the height) of 3 cm.
The auxiliary line will bisect PL, as point O is equidistant
from P and L.
EXAMPLE 3
EXAMPLE 3
a2
+ b2
= c2
EXAMPLE 3
a2
+ b2
= c2
42
+ 32
= (PO)2
EXAMPLE 3
a2
+ b2
= c2
42
+ 32
= (PO)2
16 + 9 = (PO)2
EXAMPLE 3
a2
+ b2
= c2
42
+ 32
= (PO)2
16 + 9 = (PO)2
25 = (PO)2
EXAMPLE 3
a2
+ b2
= c2
42
+ 32
= (PO)2
16 + 9 = (PO)2
25 = (PO)2
25 = (PO)2
EXAMPLE 3
a2
+ b2
= c2
42
+ 32
= (PO)2
16 + 9 = (PO)2
25 = (PO)2
25 = (PO)2
PO = 5 cm

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Geometry Section 4-4

  • 1. SECTION 4-4 Proving Triangles Congruent: ASA, AAS
  • 2. ESSENTIAL QUESTIONS How do you use the ASA Postulate to test for triangle congruence? How do you use the AAS Postulate to test for triangle congruence?
  • 3. VOCABULARY 1. Included Side: Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
  • 4. VOCABULARY 1. Included Side: The side between two consecutive angles in a triangle Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
  • 5. VOCABULARY 1. Included Side: The side between two consecutive angles in a triangle Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: If two angles and the included side of one triangle are congruent to two angles and included side of a second triangle, then the triangles are congruent Theorem 4.5 - Angle-Angle-Side (AAS) Congruence:
  • 6. VOCABULARY 1. Included Side: The side between two consecutive angles in a triangle Postulate 4.3 - Angle-Side-Angle (ASA) Congruence: If two angles and the included side of one triangle are congruent to two angles and included side of a second triangle, then the triangles are congruent Theorem 4.5 - Angle-Angle-Side (AAS) Congruence: If two angles and the nonincluded side of one triangle are congruent to the corresponding angles and nonincluded side of a second triangle, then the triangles are congruent
  • 7. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL Given: L is the midpoint of WE,WR ! ED
  • 8. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL Given: L is the midpoint of WE,WR ! ED 1. L is the midpoint of WE,WR ! ED
  • 9. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 1. L is the midpoint of WE,WR ! ED
  • 10. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 1. L is the midpoint of WE,WR ! ED
  • 11. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 1. L is the midpoint of WE,WR ! ED
  • 12. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 1. L is the midpoint of WE,WR ! ED
  • 13. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm. 1. L is the midpoint of WE,WR ! ED
  • 14. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm. 4. ∠LWR ≅ ∠LED 1. L is the midpoint of WE,WR ! ED
  • 15. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm. 4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm 1. L is the midpoint of WE,WR ! ED
  • 16. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm. 4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm 5. △WRL ≅△EDL 1. L is the midpoint of WE,WR ! ED
  • 17. EXAMPLE 1 Prove the following. Prove: △WRL ≅△EDL 1. Given Given: L is the midpoint of WE,WR ! ED 2. WL ≅ EL 2. Midpoint Thm. 3. ∠WLR ≅ ∠ELD 3. Vertical Angles Thm. 4. ∠LWR ≅ ∠LED 4. Alternate Interior Angles Thm 5. △WRL ≅△EDL 5. ASA 1. L is the midpoint of WE,WR ! ED
  • 18. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM
  • 19. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM
  • 20. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given
  • 21. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N
  • 22. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N 2. Reflexive
  • 23. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N 2. Reflexive 3. △JNM ≅△KNL
  • 24. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N 2. Reflexive 3. △JNM ≅△KNL 3. AAS
  • 25. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N 2. Reflexive 3. △JNM ≅△KNL 3. AAS 4. LN ≅ MN
  • 26. EXAMPLE 2 Prove the following. Prove: LN ≅ MN Given: ∠NKL ≅ ∠NJM, KL ≅ JM 1. ∠NKL ≅ ∠NJM, KL ≅ JM 1. Given 2. ∠N ≅ ∠N 2. Reflexive 3. △JNM ≅△KNL 3. AAS 4. LN ≅ MN 4. Corresponding Parts of Congruent Triangles are Congruent (CPCTC)
  • 27. EXAMPLE 3 On a template design for a certain envelope, the top and bottom flaps are isosceles triangles with congruent bases and base angles. If EV = 8 cm and the height of the isosceles triangle is 3 cm, find PO.
  • 28. EXAMPLE 3 On a template design for a certain envelope, the top and bottom flaps are isosceles triangles with congruent bases and base angles. If EV = 8 cm and the height of the isosceles triangle is 3 cm, find PO. EV ≅ PL, so each segment has a measure of 8 cm. If an auxiliary line is drawn from point O perpendicular to PL, you will have a right triangle formed.
  • 29. EXAMPLE 3 On a template design for a certain envelope, the top and bottom flaps are isosceles triangles with congruent bases and base angles. If EV = 8 cm and the height of the isosceles triangle is 3 cm, find PO. EV ≅ PL, so each segment has a measure of 8 cm. If an auxiliary line is drawn from point O perpendicular to PL, you will have a right triangle formed. In the right triangle, we have one leg (the height) of 3 cm. The auxiliary line will bisect PL, as point O is equidistant from P and L.
  • 32. EXAMPLE 3 a2 + b2 = c2 42 + 32 = (PO)2
  • 33. EXAMPLE 3 a2 + b2 = c2 42 + 32 = (PO)2 16 + 9 = (PO)2
  • 34. EXAMPLE 3 a2 + b2 = c2 42 + 32 = (PO)2 16 + 9 = (PO)2 25 = (PO)2
  • 35. EXAMPLE 3 a2 + b2 = c2 42 + 32 = (PO)2 16 + 9 = (PO)2 25 = (PO)2 25 = (PO)2
  • 36. EXAMPLE 3 a2 + b2 = c2 42 + 32 = (PO)2 16 + 9 = (PO)2 25 = (PO)2 25 = (PO)2 PO = 5 cm