- Ellie Krumsieg
Proposition 1.10 - Bisect a given finite straight line.
Proof Let AB be a given finite straight line. By proposition 1.1, construct an equilateral triangle ABC. By proposition 1.9, Let <ACB be bisected by CD such that D is the intersection of CD and AB. Now, consider triangles ACD and CDB. AC=CB, <ACD=<DCB, and CD=DC. So, by SAS, triangles ACD=DCB. Thus, AD=DB by similar triangles.