We have two different triangles $\triangle ABC$ and $\triangle A'B'C'$. Now we have a relation between heights $AH$ and $A'H'$.Actually $H$ and $H'$ are the points on $BC$ and $B'C'$ sides. $$\frac{AH}{A'H'}=x$$ Now we must prove that $$ \triangle ABC \thicksim \triangle A'B'C'$$ Actually I have done something. But unhopefully no results! If $\triangle ABC$ and $\triangle A'B'C'$ are similiar then we have: $$\frac{AC}{A'C'}= \frac{AB}{A'B'}= \frac{BC}{'BC'}$$ But it must be proved and we don't know that it is true or not.
And $$AH=AC\sin\angle C$$ $$A'H'=A'C'\sin\angle C'$$ Now any ideas to prove that these triangles are similiar?!
It doesn't make sense, to me, saying that one pair of anything is proportional. So I take for granted the OP meant that the three pairs of attitudes are proportional, that is $$\frac{h_a}{h^\prime_a} = \frac{h_b}{h^\prime_b} = \frac{h_c}{h^\prime_c}.$$
In this case, it is easy to imply that the two triangles are similar, because $h_a,h_b,h_c$ are reversely proportional to $a,b,c$. More precisely $$\frac{h_a}{h_a^\prime} = \frac{2S/a}{2S^\prime/a^\prime} = \frac{a^\prime}{a} \cdot \color{red}{\frac S{S^\prime}},$$ and so $$\frac{a^\prime}a = \frac{b^\prime}{b} = \frac{c^\prime}c.$$