∴S△ECF:S△ACB= 1:2
You: EF AB
∴△ECF∽△ACB
s△ecf:s△acb=(ce/ca)^2= 1/2
And AC = 4.
∴CE=2√2
(2) let the length of CE be x.
∫△ECF∽△ACB
∴CE/CA=CF/CB
∴CF=3/4x
According to the fact that the perimeter of △ECF is equal to the perimeter of quadrilateral EABF, it can be concluded that
x+CE+3/4X=(4-X)+5+(3-3/4X)+CE
The solution is all-weather
The length of ce is 24/7.