Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one. (Proposition 20)
[Desmos step-by-step graphs can be found here]
Let three lines A, B, and C satisfy the condition that any two taken together are greater than the remaining one.
We will now construct a triangle whose sides are equal to these three lines.
Set out a straight line beginning at a point D that continues through point E.
Place a point F such that DF is equal to A, a second point G such that FG is equal to B, and a third point H such that GH is equal to C.
(Proposition 3 allows us to create a line segment of length equal to the length of a given line segment.)
Construct a circle centered at F with a radius equal to DF. This circle can be labeled as DKL.
Construct a second circle centered at G with a radius equal to GH. This circle can be labeled as HKL, where K and L are the two points of intersection between the two circles.
Let FK and GK be joined.
We will now show that triangle FGK has three sides that are equal to A, B, and C.
- Since K is on the circle centered at F, FK is equal to the radius DF which is also equal to A. Therefore FK is equal to A.
- Since K is also on the circle centered at G, GK is equal to the radius GH which is also equal to C. Therefore GK is equal to C.
- FG was set up to be equal to B.
FK, FG, and GK are equal to the three straight lines A, B, and C.
Therefore a triangle with sides equal to the three lines has been constructed.
Does triangle FGL also satisfy the requirements? As an exercise, see if you can show that triangle FGL has sides that are the lengths of A, B, and C.
The next proposition is another construction, requiring us to construct an angle that is equal to a given angle. The construction in Proposition 22 plays a large role in that construction.
George