Find if a list is an ABC-triple
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

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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.