Six Lines, Six Sides, One Point

$a+b = c+e \quad (1)$

$a+c = e \quad (2)$

$a+e = c+d \quad (3)$

$2a = e+f \quad (4)$

From the first and second constraint, we can conclude: $b = 2c$
From the second and third constraint, we can conclude: $d = 2(e-c) = 2a$

Then according to Ceva's theorem , $\dfrac{a}{b}\cdot\dfrac{c}{d}\cdot\dfrac{e}{f} =1$ .

Thus, $\dfrac{a}{2c}\cdot\dfrac{c}{2a}\cdot\dfrac{e}{f} =1$ .
Hence, $e = 4f$ .

Referring to the fourth constraint, $2a = 5f$ ,

Thus, $a = 5k$ and $f = 2k$ for some positive integer $k$ .

Then $e = 4f = 8k$ .

$d = 2a =10k$ .

$c = e - a = 8k - 5k = 3k$ .

Finally, $b = 2c = 6k$ .

Then $a+b = 11k$ ; $c+d = 13k$ ; $e+f = 10k$ .

Since all the three sides of this triangle are co-prime, $k=1$ .

As a result, the perimeter equals $a+b+c+d+e+f = 5+6+3+10+8+2 = \boxed{34}$ .

None of the constants is 0.

a+b = c+e ........(1)
a+c = e ........... (2)
a+e = c+d ...... (3)
2a = e+f ......... (4)

(1) - (2)...gives ......................b=2c ...........(5)
(2) + (3).gives ......................d=2a ............(6)

By Ceva's Theorem,
a * c * e = b * d * f
From (5) and (6),
a * c * e =2c * 2d * f
So e = 4f, By (4) e + f = even.
So likely ................f = 2, ...e = 8.

From (4) ................a = 5.
From (2) ................c = 3.
From (5) ................b = 6.
From (6) ................b = 10.

The sides are ,
5+6=11,......3+10=13,........8+2=10.
They are co-prime and integers.

So perimeter = 11 + 13 + 10 = $\color{#D61F06}{ \huge 34}$