If two obtuse triangles inscribed in a circle, are the triangles similar? 
Suppose I inscribe an obtuse triangle DEF in the circle, triangle DEF has the length a and b as well. Is triangle ABC and DEF similar?
 A: Yes, they are congruent.
By the Inscribed Angle Theorem, if the center of the circle were to lie within one of the inscribed triangles, then the triangle in question would immediately be acute, as each angle would be half of a central angle which is less than 180 degrees. Similarly, if the center were to lie on an edge of one of the inscribed triangles, then the triangle in question would immediately be a right triangle whose hypotenuse is a diameter of the circle. Therefore, the center of the circle lies outside both of the inscribed triangles, and the obtuse angle is at the vertex opposite the edge nearest to the center (that is, at $B$ and $E$, in your first picture).
To avoid confusion, I'll stick with your first picture, supposing that $a=d,$ that $c=f,$ and that $r$ is the radius of the circle. To show that the triangles are congruent, it suffices by SAS to show that $m\angle B=m\angle E.$
By the Law of Sines, $$m\angle A=\sin^{-1}\left(\frac{a}{2r}\right)=\sin^{-1}\left(\frac{d}{2r}\right)=m\angle D$$ and $$m\angle C=\sin^{-1}\left(\frac{c}{2r}\right)=\sin^{-1}\left(\frac{f}{2r}\right)=m\angle F.$$ Since the interior angles of a triangle sum to two right angles, it follows that $m\angle B=m\angle E,$ as desired.
