The argument about conics through the three vertices of a triangle A, B, C and an interior point goes like this. One needs to know about involutions, that is, homographies of order 2. In real geometry, homographies come in 3 types: hyperbolic, which have two distinct (real) fixed points; parabolic, where the fixed points coincide (and are real); and elliptic, where there are no real fixed points, i.e., the fixed points are complex conjugate. For an involution, the parabolic case doesn't occur; in the hyperbolic case the pairs of associated points, (A, A'); (B, B') etc are either nested (like A, B, B', A') or separated (like A, A', B, B'), whereas for an elliptic involution they separate each other, like A, B, A', B'.

Now, given four points, you look at the pencil of conics through these points. The conics of the pencil cut point-pairs of an involution on any line, including the line at infinity.

Let S_{i} := a_{i}x^{2} + 2h_{i}xy + ... = 0, i=1,2, be two conics of the pencil through A, B, C, D, so that the
general member of the pencil is S(lambda) := S_{1}+lambda.S_{2} = 0.
Put f(lambda) = (a_{1}+lambda.a_{2})(b_{1}+lambda.b_{2})-(h_{1}+lambda.h_{2})^{2}, a
quadratic in lambda that is positive for an ellipse
and negative for a hyperbola. It will always have at least one
negative value, since the pencil always has at least one
non-parallel line-pair. So it has positive values (and hence the
pencil contains ellipses) iff it has distinct real zeros; and then it has parabolas also. If it doesn't contain positive values, all the proper conics in the pencil are hyperbolas. Confusingly, this means that the pencil consisting only of hyperbolas corresponds to an elliptic involution cut on the line at infinity.

Now you finish the argument by looking at A, B, C, D with D inside triangle ABC. It is clear (I hope) that the line pair AD, CD cut a pair of points on the line at infinity that separates the points cut by the line pair BD, AC. Thus the involution is elliptic, and all the proper conics of the pencil are hyperbolas. (And, if no one of A, B, C, D is inside the triangle formed by the other three, the corresponding pairs of points cut by AD, CD and by BD, AC on the line at infinity do not separate each other, so the involution is hyperbolic, and the pencil then contains ellipses, hyperbolas, and just two parabolas, though either or both of those might degenerate into parallel line pairs - for example if ABCD is a trapezium, or a parallelogram.)