Let $ABC$ and $A'B'C'$ be the two triangles inscribed in the same conic.
Consider the exagon $ABCC'B'A'$: by Pascal's theorem the three points
AB&\cap C'B'& BC&\cap B'A'& CC'&\cap A'A&
lie on a straight line.
Now, let $a=BC$, $b=AC$, $c=AB$ and $a'=B'C'$, $b'=A'C'$, $c'=A'B'$ be the edges of our triangles and consider the exagon whose edges are $a,b,c,a',b',c'$.
The three lines containing the three pairs of opposite vertex are
(a\cap b)(a'\cap b')&=CC'\\
(b\cap c)(b'\cap c')&=AA'\\
(c\cap a')(c'\cap a)&=(AB\cap B'C')(A'B'\cap BC)
and they meets at a point.
By Brianchon's theorem $a,b,c,a',b',c'$ are tangent to a conic.