Prove that $(a,c)\cdot (a,d)=a\cdot(a,(b,(c,d)))$ if $a,b,c,d\in\mathbb{Z}$ and $ab=cd$

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Use the property $(ab,ac)=a\cdot (b,c)$ - the greatest common factor feature

Then $a\cdot(a,(b,(c,d))) = (a^2,a\cdot(b,(c,d))) = (a^2,(ab,a\cdot(c,d))) = (a^2,(ab,(ac,ad))) = (a^2,(cd,(ac,ad)))$

Is this correct? How to get $(a,c)\cdot (a,d)$?

I got!! There is the right solution:

We know that $(ab,ac)=a\cdot (b,c)$ and we know gcd's associativity and commutativity.

Then $a(a,(b,(c,d))) = (a^2,a(b,(c,d))) = (a^2,(ab,a(c,d))) = (a^2,(ab,(ac,ad))) = (a^2,(cd,(ac,ad))) = (a^2,(cd,(ad,ac))) = (a^2,((cd,ad),ac)) = (a^2,(d(c,a),ac)) = (a^2,(ac,d(c,a))) = ((a^2,ac),d(c,a)) = (a(a,c),d(a,c)) = (a,c)\cdot (a,d)$

Thank You.