Math

Place the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 in the circles
so that all three sides of the triangle have the same sum.

The numbers 1-9 add up to 45.
The sum of all three sides = 45 + (3 corners),
since each corner contributes to two sides.
Sum of all three sides = 3 * one side.
(3 corners) = sum of all three sides - 45.
Since both terms on the right side are divisible by 3,
the sum of the three corners is divisible by 3 too.
That limits the choices for the corners.
The sum of each side then will be (45 + (3 corners))/3.
From there, exhaustive search reveals the solutions below.

Each solution has a complement, where each number, k, is replaced by 10-k.
In the first two rows, the complements are one above the other,
In the bottom row, they are all self-complementary, with a reflection about the vertical axis.
The last pair are the only ones where, for a given choice for the corners, there is only one solution. The others are all in pairs.

Solutions

Side = 17			Side = 19			Side = 19
2 5   4 9     8 1  6 7  3	2 6   5 8     7 1  4 9  3		4 5   2 9     6 1  3 8  7	4 6   3 8     5 1  2 9  7		3 5   1 9     8 2  4 6  7	3 6   4 8     5 2  1 9  7
Side = 23			Side = 21			Side = 21
8 5   6 1     2 9  4 3  7	8 4   5 2     3 9  6 1  7		6 5   8 1     4 9  7 2  3	6 4   7 2     5 9  8 1  3		7 5   9 1     2 8  6 4  3	7 4   6 2     5 8  9 1  3
Side = 20			Side = 20			Side = 20	Side = 20
5 9   1 2     8 4  3 7  6	5 3   7 8     2 4  1 9  6		5 4   6 9     1 2  3 7  8	5 6   4 7     3 2  1 9  8		5 6   4 8     2 1  3 7  9	5 4   6 8     2 3  1 9  7