Sss geometry example tricia cole12/13/2023 I say that the angle BAC is also equal to the angle EDF.įor, if the triangle ABC be applied to the triangle DEF, and if the point B be placed on the point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF. Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF and let them have the base BC equal to the base EF. On this page I have collected a few proofs of the SSS criterion from several popular geometry texts. But he took care to make the two formulations sound alike so as to draw attention to the relationship between the two that he built and exploited. He did not find it necessary to mention that fact in a later proposition. Further, in Elements I.4, Euclid indeed concludes that under the SAS conditions, the remaining side and the remaining two angles in one triangle equal their counterparts in the other. As you'll see shortly, Euclid has based his proof of SSS on that of SAS. The reason for a strange, asymmetric formulation is that Elements I.4 is about what is known nowadays as SAS and establishes the congruence of two triangles by a pair of equal sides and the angle they include. Elements I.8 reads, If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines. In Euclid's Elements the formulation is somewhat different. Thus, the two triangles (∆ABC and ∆DEF) are congruent by the SAS criterion.The acronym SSS (side-side-side) refers to the criterion of congruence of two triangles: if the three sides of one equal the three sides of the other, the two triangles are congruent. 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.
0 Comments
Leave a Reply.AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |