If $\frac{AB}{A'B'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$ then prove that $\triangle{ABC}$ is similar to $\triangle{A'B'C'}$
How to prove that angles are equal? I don't really know where to start
Edit : solution shouldn't use trigonometry.
Edit : Similarity definition: when corresponding angles of two shapes are equal and corresponding sides are proportional.
If you want to prove that the three angles are orderly congruent:
Consider the triangle $\triangle ABC$ and the half-lines $Ab:=\overrightarrow{AB},\,Ac:=\overrightarrow{AC}$ which border the angle $\angle BAC$. There is exactly one point $B''\in Ab$ such that $A'B'\cong AB''$ and exactly one point $C''\in Ac$ such that $A'C'=AC''$.
By the inverse of Thales' intercept theorem, $B''C''\parallel BC$. Hence, $\angle C''B''A\cong \angle CBA$ and $\angle B''C''A\cong \angle BCA$ by parallel theorem.
By Thales' intercept theorem, $BC:B''C''=AB:AB''$ and, since $AB''\cong A'B'$, this implies that $B''C''\cong B'C'$.
Ny the third congruence theorem (SSS), $\triangle AB''C''\cong \triangle A'B'C'$ and, hence, $\angle BAC\cong\angle B'A'C'$, $\angle ABC\cong \angle AB''C''\cong A'B'C'$ and $\angle ACB\cong\angle AC''B''\cong \angle A'C'B'$.
Q.E.D.