Let $l$ be the sum of each of the $5$ lines. Adding up the $4$ square sides you get: $4l=123\times 2+15+12+6+9$ then $l=72$ Let $x$ the value on the rightmost vertex. Then adding the left triangle: $3l=2\times (123-x)+12+15+17$ then $216+2x=290$ then $x=37$ So finally the numbers are $37, 29, 31,26$

Let the the intersection of the sides with the numbers $15$ and $9$ be $ A$ and the other circles on the clockwise, respectively $B, C, D$. $A+B+9 = B+C+6 = D+C+12 = D+A+15 = A+C+17 = S$. $2S = A+B+9+C+D+12 = 144$, once that $A+B+C+D = 123$. This give us that $S = 72$. $(A+B+9)+(A+D+15)+(A+C+17) =2A + 123 + 41= 3S = 216.$ So $A = 26$. $A+B+9 = S=72$ give us $B = 37$. $B+C+6 = S = 72 \Rightarrow C = 29$. $A+D+15 = S = 72 \Rightarrow D = 31$. So the answer is $(A,B,C,D) = (26,37,29,31)$(sorry for any latex mistake, I'm still learning it)