Prove that $\gcd(a+b,\text{lcm}(a,b))=\gcd(a,b)$ where $a,b\in\mathbb{Z}$. [duplicate]

Show that if $a$, $b$ are positive integers, then we have: $\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.

Below are a few proofs. First, here's a couple from my sci.math post on 2001/11/10
For variety here is yet another using $\rm\ (a,b)\ [a,b]\ =\ \color{#c00}{ab}\ \ $ and basic gcd laws:

$\rm\quad (a,b)\ (a\!+\!b,\, [a,b])\, =\, (aa\!+\!ab,\, ab\!+\!bb,\, \color{#c00}{ab})\ =\ (aa,\,bb,\,ab)\, =\, (a,b)^2$

By the way, recall that the key identity in the second proof arose the other day in our discussion of Stieltjes $\rm\ 4\:n+3\ $ generalization of Euclid's proof of infinitely many primes. Here's a slicker proof:

Lemma $\rm\ \ (a\!+\!b,\,ab) = 1 \iff (a,b) = 1$

Proof $\rm\ \ \ (\color{#90f}a,b)^2 \color{#C00}{\large\subseteq} (a\!+\!b,\,ab) \color{#0a0}{\large \subseteq} (a,b)\ \ $ since, $ $ e.g. $\rm\ \ \color{#90f}{a^2} = a(a\!+\!b)-ab\color{#c00}{\large \in} (a\!+\!b,\,ab)$.

$\rm\ 1\in (a\!+\!b,\, ab) \color{#0a0}{\large\subseteq}(a,b)\Rightarrow 1\in (a,b).\:$ Conversely $\rm\ 1 \in (a,b) \Rightarrow 1 \in (a,b)^2\color{#c00}{\large\color{#c00}\subseteq} (a\!+\!b,\,ab)$

Another Dubuquesque attempt; for legibility, write $d=\gcd(a,b)$: \begin{align*} \gcd\Bigl(d(a+b), ab\Bigr) &= \gcd\Bigl(d(a+b), ab, ab\Bigr)\\ &=\gcd\Bigl(d(a+b),\ ab-a(a+b),\ ab-b(a+b)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ a^2,\ b^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a^2,b^2)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a,b)^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ d^2\Bigr)\\ &= d\gcd\Bigl(a+b,d\Bigr)\\ &= d\gcd\Bigl(a+b,\gcd(a,b)\Bigr)\\ &= d\gcd(a,b)\\ &= \gcd(a,b)\gcd(a,b). \end{align*}

(Second line uses the fact that $a(a+b)$ and $b(a+b)$ are both multiples of $d(a+b)$).

Now divide through by $\gcd(a,b)$ to get the desired result.