Post by karl on Mar 2, 2020 18:42:56 GMT
Hello,
I'm not sure but I propose the Following demonstration:
1) If ABCD is a rectangle, it is meaning that each edge at right angle.
2) To prove it, I propose to prove that the each opposed edge on the unity circle pass are placed on a line passing through the center.
3) For example, if A=-C and B=-D => (A,C) and (B,D) are on a stright line through the center and thanks to the Thales theorem, A, B, C and D have right angle.
4) If ABCD is a rectangle:
(a) A=-C
(b) B=-D
(a) + (b) A+B = -(C+D) <=> A+B+C+D=0
Done!
(It is my first time that I'm trying to demonstrate somethink)
I'm not sure but I propose the Following demonstration:
1) If ABCD is a rectangle, it is meaning that each edge at right angle.
2) To prove it, I propose to prove that the each opposed edge on the unity circle pass are placed on a line passing through the center.
3) For example, if A=-C and B=-D => (A,C) and (B,D) are on a stright line through the center and thanks to the Thales theorem, A, B, C and D have right angle.
4) If ABCD is a rectangle:
(a) A=-C
(b) B=-D
(a) + (b) A+B = -(C+D) <=> A+B+C+D=0
Done!
(It is my first time that I'm trying to demonstrate somethink)
