Similarity by SSA (side-side-angle) in obtuse triangles Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54:

&quot;In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other acute angles are proportional. Prove that the triangles are similar.&quot;

I tried this first by cosine rule, and the question reduced to:
Assume $\triangle ABC$ and $\triangle DEF$ with $\angle A$ and $\angle D$ being obtuse, $\angle B=\angle E$ and $AC/BC=DF/EF$. We may take $DF=k\cdot AC$, $EF=k\cdot BC$ and apply $\cos B= \cos E$, obtaining $DE=k\cdot AB$ and the problem is solved. EDIT: Even this however is difficult to prove, as pointed out by cosmo5 in his answer.
What is interesting and challenging is to prove the same without the cosine law (probably the expected solution, as cosine law doesn't seem to help), as the law hasn't been introduced in the book until later chapters, while this problem is taken from chapter 3.
My attempt:
Fix $\triangle ABC$ as well as points $E$ and $F$ of $\triangle DEF$. Assuming $DF=k\cdot AC$, $EF=k\cdot BC$, the locus of $D$ will be a circle with centre $F$ and radius $k\cdot AC$.
After this however, I couldn't go further with the same concept. Either a hint or a solution related to my thought process, or any other thought process for that matter, would be greatly appreciated.
Hint: note that there are at most two points on ray $BA$ which is of a certain distance from point $C$ (because you are intersecting a ray with a circle).
They are congruent if the known congruent angles are obtuse. If the known congruent angles are acute, then the triangles are congruent if the triangles are both acute or both obtuse.  Why this is not more widely distributed is above my paygrade!