Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. :)

triangle

share|improve this question

2 Answers 2

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

share|improve this answer

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 $\overline{AB}\parallel\overline{C'A'}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.