Jtdylktuy

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
$$AH=ACsinangle C$$
$$A'H'=A'C'sinangle C'$$
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
$$frac{a^prime}a = frac{b^prime}{b} = frac{c^prime}c.$$

It does not imply that triangles whose heights are proportional are necessarely similar. In fact, every pair of triangles have proportional heights. There is probably something missing in your problem.