As $(ABC'S), (BCA'S), (CAB'S)$ are vertexes of $3$ isosceles trapezoids, points $A', B', C'$ lie on intersection of parallel plane to $ABC$ through $S$ with sphere $SABC;$ also $SA'||BC, SB'||AC, SC'||AB.$ Let $a, b, c$ be the lines of intersection of $ABC$ with $AB'C',$ $A'BC'$ and $A'B'C;$ obviously $B'C'||a, A'C'||b, A'B'||c.$ We need to prove, that $a, b, c$ are concurrent. Let $S'$ be the second point of intersection of $a$ with circle $(ABC).$ From all parallel lines we have $$\angle (S'C, CB) = \angle (S'A, AB) = \angle (a, AB) = \angle (B'C', C'S) = \angle (B'A', A'S) = \angle (c, CB),$$i.e. $S'$ lies on $c;$ analogously it lies on $b,$ that ends our proof.

This is exactly how i solved it during the contest

I didn't solve my 9.6 (((

Interestingly, it seems that one needs only the $SA' \parallel BC$, $SB' \parallel AC$ and $SC' \parallel AB$ condition, but the fact that the trapezoids are isosceles is redundant. This is based on the following simple observation: if three lines, which are parallel to the sides of the first triangle, and are passing through vertices of the second triangle, are concurrent, then three lines, which are parallel to the sides of the second triangle, and are passing through vertices of the first triangle, are concurrent, too. So the condition is actually overloaded.