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

"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."

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.
 A: 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).
A: The conditions given are essentially $SSA$ but there is no such criterion for congruence or similarity. This is because exactly two triangles are possible as shown.

Both $\triangle DEF$ and $\triangle D'EF$ satisfy the conditions that sides about second acute angle are proportional and first acute angles are equal. But both cannot be similar to $\triangle ABC$ at the same time.
Trigonometry cannot help either. For example, by cosine-rule
$$\cos B=\cos E=\frac{x^2+(ka)^2-(kb)^2}{2\cdot x\cdot ka}$$
is a quadratic in $x$; it will produce two solutions namely $DE$ and $D'E$ for the third side.
The extra information that the third angle is obtuse pins down the unique triangle $DEF \sim \triangle ABC$. Since $D,D'$ must lie on either side of perpendicular dropped from $F$ to $BC$, one of the two angles $(\angle ED'F)$ is acute while other $(\angle EDF)$ is obtuse.
I think the book is expecting this kind of analysis as answer. If you're interested, see my answer to this problem. A user with a keen eye spots that the solution can be kept elementary, exactly by using the above condition that if two triangles match by $SSA$ and it is known that both have obtuse angles, then they are congruent/similar.
A: 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!
