Don't touch the center. It is already even.
Start anywhere by connecting a dotted line from one vertex to the next. To keep things so we know what we are talking about, go clockwise. Now you have 2 points that are Eulerized that were not before.
Skip and edge and do the same thing to the next two vertices. Those two become eulerized. Skip an edge and do the last 2.
Let's try to describe this better. Start at any vertex and number them 1 to 6 clockwise.
Join 1 to 2
Join 3 to 4
Join 5 to 6
I think 3 is the minimum.
3 <<<< answer
9:12, 18:24,6:8
3*3=9 and 4*3=12
3*6=18 and 4*6=24
3*2=6 and 4*2=8
Given:
ABC is an isosceles triangle in which AC =BC.
D and E are points on BC and AC such that CE=CD.
To prove:
Triangle ACD and BCE are congruent.
Solution:
In triangle ACD and BCE,
(Given)

(Common angle)
(Given)

In triangles ACD and BCE two corresponding sides and one included angle are congruent. So, the triangles are congruent by SAS congruence postulate.
(SAS congruence postulate)
Hence proved.