Ninth Grade Mathematics: SAS and ASA Congruence Criteria
To prove that two triangles are congruent, mathematicians do not need to measure all six components. Instead, specific combinations of three known parts establish total congruence. In this chapter, you will master the two foundational criteria that involve included parts: Side-Angle-Side (SAS) and Angle-Side-Angle (ASA).
The SAS (Side-Angle-Side) Congruence Criterion
The Side-Angle-Side criterion is the primary axiom of triangle congruence from which other criteria are derived.
The SAS Congruence Criterion:
Two triangles are congruent if two sides and the INCLUDED angle of one triangle
are equal to two sides and the INCLUDED angle of the other triangle.
TRIANGLE ABC TRIANGLE DEF
A D
* *
/ \ / \
b / \ c b / \ c
/ A \ / D \
*-------* *-------*
B C E F
The word included is mathematically critical. The angle must be the one formed between the two known sides (the angle trapped between them). If the equal angle is not between the two known sides (giving SSA), congruence is not guaranteed.
Valid SAS:
Side AB = Side DE
Included Angle A = Included Angle D
Side AC = Side DF
Conclusion: Triangle ABC =~ Triangle DEF
Why SSA (Side-Side-Angle) Fails
If the known angle is not the included angle, the geometric configuration is ambiguous. An arc swung from the free vertex can intersect the opposite side in two completely different locations, creating two non-congruent triangles with the exact same two sides and non-included angle:
The Ambiguous SSA Case:
A
*
/ \
/ \ Two different triangles (ABC1 and ABC2)
/ \ share side AB, side AC, and Angle B!
/ \
-------*---------*-------*-------
B C1 C2
Because SSA can generate two different triangles, SSA is not a valid congruence criterion.
The ASA (Angle-Side-Angle) Congruence Criterion
The Angle-Side-Angle criterion applies when two angles and the side connecting their vertices are known:
The ASA Congruence Criterion:
Two triangles are congruent if two angles and the INCLUDED side of one triangle
are equal to two angles and the INCLUDED side of the other triangle.
TRIANGLE ABC TRIANGLE DEF
A D
* *
/ \ / \
/ \ / \
/ \ / \
*-------* *-------*
B a C E a F
Angle B = Angle E Side BC = Side EF (Included)
Angle C = Angle F
The included side is the segment that joins the vertices of the two known angles.
Proof of the ASA Criterion
Unlike SAS, which is accepted as an axiom, ASA can be formally proven using SAS and proof by contradiction:
Given: Triangle ABC and Triangle DEF such that:
Angle B = Angle E, Angle C = Angle F, and side BC = EF.
To Prove: Triangle ABC =~ Triangle DEF.
Proof:
Case 1: Assume AB = DE.
Then in Triangle ABC and Triangle DEF:
AB = DE (Side), Angle B = Angle E (Angle), BC = EF (Side).
By the SAS Axiom, Triangle ABC =~ Triangle DEF.
Case 2: Assume AB != DE. Suppose AB > DE.
Take a point P on AB such that PB = DE. Join PC.
In Triangle PBC and Triangle DEF:
PB = DE, Angle B = Angle E, BC = EF.
By SAS, Triangle PBC =~ Triangle DEF.
By CPCTC, Angle PCB = Angle DFE.
However, we are given that Angle ACB = Angle DFE.
Therefore, Angle PCB = Angle ACB, which means a part equals the whole!
This is impossible unless ray PC coincides with ray AC (point P coincides with A).
Therefore, AB must equal DE, which proves Triangle ABC =~ Triangle DEF.
Chapter Exercises
Exercise 1: In quadrilateral ABCD, diagonal AC bisects Angle A, and side AB = AD. Prove that Triangle ABC =~ Triangle ADC, and deduce that side BC = CD.
Exercise 2: Line segment AB is parallel to line segment CD, and O is the midpoint of segment AD. Prove that Triangle AOB =~ Triangle DOC using the ASA criterion, and deduce that O is also the midpoint of segment BC.
Exercise 3: Explain in paragraph form why Side-Side-Angle (SSA) is not an accepted congruence criterion, sketching an ASCII diagram that illustrates the ambiguous case.
Solutions and Explanations
Solution 1: Consider Triangle ABC and Triangle ADC. We are given that side AB = side AD. Because diagonal AC bisects Angle A, Angle BAC = Angle DAC. The line segment AC is common to both triangles (AC = AC). In Triangle ABC and Triangle ADC, we have AB = AD (side), included Angle BAC = included Angle DAC (angle), and AC = AC (side). By the SAS Congruence Criterion, Triangle ABC =~ Triangle ADC. By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the corresponding third sides must be equal: BC = CD.
Solution 2: Since AB is parallel to CD and AD acts as a transversal line, the alternate interior angles are equal: Angle OAB = Angle ODC. Since AB is parallel to CD and BC acts as another transversal, Angle OBA = Angle OCD (or vertically opposite angles: Angle AOB = Angle DOC). Furthermore, we are given that O is the midpoint of AD, so side OA = side OD. In Triangle AOB and Triangle DOC: Angle OAB = Angle ODC (angle), side OA = side OD (included side), and Angle AOB = Angle DOC (angle). By the ASA Congruence Criterion, Triangle AOB =~ Triangle DOC. By CPCTC, side OB = side OC, which proves that point O is also the midpoint of segment BC.
Solution 3: Side-Side-Angle (SSA) is not a valid congruence criterion because knowing two side lengths and a non-included angle does not determine a unique triangle. If the side opposite the given acute angle is shorter than the adjacent side but longer than the perpendicular altitude, a compass swung from the top vertex can intersect the opposite baseline in two distinct points:
A
*
/|\
/ | \
c / | \ a
/ |h \
/ | \
*-----+-----*-------*
B C1 C2
Both Triangle ABC1 and Triangle ABC2 share the exact same side c (AB), the exact same side a (AC1 = AC2), and the exact same non-included acute Angle B. Yet Triangle ABC1 is an acute triangle while Triangle ABC2 is an obtuse triangle, and they are clearly not congruent. Because two different triangles can be formed with the same measurements, SSA fails the uniqueness requirement of geometric congruence.