From Euclidss Elements  (see here)

Common Notions (CN)
 1 Things equal to the same thing are also equal to one another 2 And if equal things are added to equal things then the wholes are equal 3 And if equal things are subtracted from equal things then the remainders are equal 4 And things coinciding with one another are equal to one another 5 And the whole [is] greater than the part

Postulates (Post)

 1 Let it have been postulated to draw a straight-line from any point to any point 2 And to produce a finite straight-line continuously in a straight-line 3 And to draw a circle with any center and radius 4 And that all right-angles are equal to one another 5 And that if a straight-line falling across two (other) straight-lines makes internal angles on the same side (of itself whose sum is) less than two right-angles, then the two (other) straight-lines, being produced to infinity, meet on that side (of the original straight-line) that the (sum of the internal angles) is less than two right-angles (and do not meet on the other side)

Proposition 1
To construct an equilateral triangle on a given finite straight-line.
Proof:
0)  Let AB be the given finite straight-line
1)  I use Post3 by considering A as center and the finite straight-line AB as radius
2)  I use Post3 again, by considering B as center and the finite straight-line AB as radius
3)  Let C be a point where the circles cut one another. I use Post1 by considering A and C as points
4)  I use Post1 again, by considering B and C as points     [click the image if you want to enlarge it]

5)  Thus: AC = AB       6)  and: BC = AB       7)  From 5) and 6) I get: AC = BC
8)  From 5), 6) and 7) I get that the triangle ABC is equilateral.

The existence of common points between straight lines is guaranteed by Post5, but, without continuity, who guarantees me the existence [asserted in 3)] of common points between "curves" that bypass

 • The existence of common points between straight lines is guaranteed by Post5, but, without continuity, who guarantees me the existence [asserted in 3)] of common points between "curves" that bypass?

Proposition 4
If two triangles have two sides equal to two sides, respectively, and have the angle(s) enclosed by the equal straight-lines equal, then they will also have the base equal to the base, and the triangle will be equal to the triangle, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles
 Proof           [click the image if you want to enlarge it] : 0)  Let ABC and DEF be two triangles having: AB=DE, AC=DF, ∠A=∠D 1)  I transport the first triangle on the second so that the point A overlaps D and the segment AB is arranged along DE 2)  Since AB=DE,by 1) it turns out that B is superimposed on E and AB is superimposed on DE 3)  Since ∠A=∠D and, for 1), AB is disposed along DE, we have that AC is disposed along DF 4)  Since AC=DF and, for 1), A superimposed on D, by 3) it turns out that C is superimposed on F and AC is superimposed on DF 5)  Since by 2) and 4) B and C are superimposed on E and F, it turns out that BC is superimposed on EF 6)  Since, by per CN4, 2), 4) and 5), the triangles ABC and DEF are equal

 • [In 1)] an operation of "transport" of figures whose possibility is not postulated is performed.    Moreover: • From the fact that there is an M1 movement carrying ∠A to ∠D, an M2 movement carrying the AB segment in the DE segment and an M3 movement carrying the AC segment in the DFsegment, I cannot deduce that there is a "movement" (M1 or another) that he does all threethings.  And then, given two points, Post1 ensures the existence of a segment that has themfor extremes, not its uniqueness.