Proof: AAS Criterion
Proof: AAS Criterion
The angle-angle-side criterion for congruent triangles says that: If two angles and a non-included side of one triangle are congruent respectively to two angles and a non-included side of another triangle, then the two triangles are congruent.
Given triangles ABC and DEF with A D, C F, and BC EF. Prove that
ABC and DEF.
Proof:
Let ABC and DEF with A D, C F, and BC EF. Prove that ABC DEF.
Note: If AC DF, then the triangles would be congruent because of SAS:
(AC DF, BC EF, C F)
Case 1: Assume AC < DF. Then by (III-1), there is a point X on DF such that AC XF. Then
ABC XEF. So A X, but this isn't the case because we assumed A D.
Thus, AC is not less than DF.
Case 2: Assume AC > DF. They by (III-1), there is a point X on AC such that XC DF.
Then XBC DEF. So X D, but this isn't the case because we assumed A D. Thus, AC is not greater than DF.
Since AC is neither less than or greater than DF, it must be equal to DF. Therefore, ABC DEF.