Find if a list is an ABC-triple Three positive integers A, B, C are ABC-triple if they are coprime,
with A &lt; B and satisfying the relation : A + B = C
Examples :

1, 8, 9 is an ABC-triple since they are coprime, 1 &lt; 8 and 1 + 8 = 9
6, 8, 14 is not because they are not coprime
7, 5, 12 is not because 7 &gt; 5

You can see this Frits Beukers 2005 presentation for more details about ABC-triples.
Input/Output
Three integers, decimal written. May be separated values or
list. Output had to be a truthy/falsy value whether the three
integers are an ABC-triple.
Note: It is important to respect integers order in the list, for example: 1, 8, 9 is not considered as the same list as 9, 1, 8 or any other combination. So first is an ABC-triple and second is not.
Thus A is the first element of the list, B the second and C the third.
Test cases
Each of the following lists should output a truthy value
[1, 8, 9]
[2, 3, 5]
[2, 6436341, 6436343]
[4, 121, 125]
[121, 48234375, 48234496]

Each of the following lists should output a falsey value
[1, 1, 2]
[1, 2, 5]
[1, 9, 8]
[4, 12872682, 12872686]
[6, 8, 14]
[7, 5, 12]
Haskell, 48 38 29 bytes

-10 bytes due to TFeld's gcd trick!
-7 bytes thanks to HPWiz for improving the co-primality test and spotting a superfluous space!
-2 bytes thanks to nimi for suggesting an infix-operator!
(a!b)c=a&lt;b&amp;&amp;a+b==c&amp;&amp;gcd a b&lt;2

Try it online!
Explanation
The first two conditions a &lt; b and a + b == c are fairly obvious, the third one uses that \$\gcd(a,b) = \gcd(a,c) = \gcd(b,c)\$:
Writing \$\gcd(a,c) = U \cdot a + V \cdot c\$ using Bézout's identity and substituting \$c = a + b\$ gives:
$$
U \cdot a + V \cdot (a + b) = (U + V) \cdot a + V \cdot b
$$
Since the \$\gcd\$ is the minimal positive solution to that identity it follows that \$\gcd(a,b) = \gcd(a,c)\$. The other case is symmetric.