If the three sides of one triangle are congruent to the three sides of another triangle, How to determine whether given triangles are congruent, and to name the postulate that is used? We must use the same rule for both the triangles that we are comparing. (This rule may sometimes be referred to as SAA).įor the ASA rule the given side must be included and for AAS rule the side given must not be included. If two angles and a non-included side of one triangle are equal to two angles and a non-included side of another triangle, then the triangles are congruent. The Angle-Angle-Side (AAS) Rule states that If two angles and the included side of one triangle are equal to two angles and included side ofĪnother triangle, then the triangles are congruent.Īn included side is the side between the two given angles. The Angle-Side-Angle (ASA) Rule states that Included Angle Non-included angle ASA Rule If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.Īn included angle is the angle formed by the two given sides. The Side-Angle-Side (SAS) rule states that If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent. The Side-Side-Side (SSS) rule states that As long as one of the rules is true, it is sufficient to prove that the two triangles are congruent. There is also another rule for right triangles called the Hypotenuse Leg rule. They are called the SSS rule, SAS rule, ASA rule and AAS rule. There are four rules to check for congruent triangles. Thus, the two triangles (∆ABC and ∆DEF) are congruent by the SAS criterion.We can tell whether two triangles are congruent without testing all the sides and all the angles of This means that our original assumption of assuming that ∠B ≠ ∠E is flawed: ∠B must be equal to ∠E. On the same segment, we cannot have two perpendiculars going in different directions. To put it even more simply, note that BX and AX should both be perpendicular to GC (why?). Is this possible? Can we have two isosceles triangles on the same base where the perpendiculars to the base are in different directions? No! What we have here is two isosceles triangles standing on the same base GC, where the perpendiculars from the vertex to the base (BX and AX) are in different directions. Similarly, since AG = AC, ∆AGC is isosceles. Now, take a good look at the following figure, in which we have highlighted to conclusions we just made (we have also marked X, the mid-point of GC): Thus, BG = BC.ĪG = DF, which is equal to AC. This leads to the following conclusions:īG = EF, which is equal to BC. Now, observe that ∆ABG will be congruent to ∆DEF, by the SAS criteria. Through B, draw BG such that ∠ABG = ∠DEF, and BG = EF, as shown below, and join A to G. One of the two angles must then be less than the other. Therefore, we begin our proof by supposing that none of the corresponding angles are equal. If we could show equality between even one pair of angles (say, ∠B = ∠E), then our proof would be complete, since the triangles would then be congruent by the SAS criterion. Consider two triangles once again, ∆ABC and ∆DEF, with the same set of lengths, as shown below: Let’s discuss the proof of the SSS criterion.
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