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Given an isosceles triangle $\triangle{BAC}$ as the one in the figure below and the reflection in line $l_{BC}$ that transforms the triangle into $\triangle{B'A'C'}$. How can I prove that $s_{AB} \parallel s_{C'A'}$? I know both sides are congruent and reflections keep angles and measures.This is the first step of a problem solution. However, it doesn't mention a theorem or an axiom so it has to be extremely easy, but I'm failing to see it.

Note: ${B=B'}$ and ${C = C'}$. I'm still learning how to use geogebra and this is also my first geometry course. :)


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AB'C and B'CA are the same angle since both are equal to ACB'.

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This is only true if $AB=AC$ (that is, if $\overline{BC}$ is the base of the isosceles triangle). If that's true, then $\angle ABC\cong\angle ACB$ by the Isosceles Triangle Theorem, and $\angle ACB\cong\angle A'CB$ because reflection preserves angle measure. You should then be able to use a theorem or postulate about parallel lines cut by a transversal to get that $\overline{AB}\parallel\overline{C'A'}$.

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