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.

  • $\begingroup$ What's your similarity definition.? $\endgroup$ – Nosrati Feb 18 '17 at 14:31
  • $\begingroup$ @MyGlasses common high-school definition of similarity as described here $\endgroup$ – user351435 Feb 18 '17 at 14:40
  • $\begingroup$ @GLASSIC The common definition of similarity of triangles in my high-school was "the sides are directly proportional" and the fact that this was equivalent to having the same angles was a theorem known as "third similarity criterion". $\endgroup$ – user228113 Feb 18 '17 at 14:52
  • $\begingroup$ @G-Sassatelli post edited and a definition for similarity is added. $\endgroup$ – user351435 Feb 18 '17 at 14:59
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    $\begingroup$ @N.S.JOHN It might not be the case, but invoking the function $\sin \theta$ to prove one of the elementary facts which put together the angles of a triangle and the ratios of their sides is very likely to produce hidden circular arguments. Not very wise, IMHO. $\endgroup$ – user228113 Feb 19 '17 at 8:38

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 $\angle ACB\cong\angle AC''B''\cong \angle A'C'B'$.


  • $\begingroup$ Add a figure if it's possible. $\endgroup$ – Nosrati Feb 18 '17 at 15:08
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    $\begingroup$ @MyGlasses I wouldn't lie if I said that I have no access to Geogebra right now, but the actual reason why I didn't do it in the first place is that I'm lazy. $\endgroup$ – user228113 Feb 18 '17 at 15:16

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