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There are two triangles. The first triangle has 3 vertices called ABC, while the second one has DEF. If vertex A = vertex D, vertex B = vertex E, and vertex C = vertex F. Is there any way that I can prove that line segment AB is equal to line segment DE?

I apologize in advance if the problem is incomprehensible, I just do not really know how to describe the problem that I am facing currently. kinda a representation of the triangle given

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  • $\begingroup$ What do you mean by vertex $A$=vertex $B$ and so on$??$ $\endgroup$
    – Vanessa
    2 days ago
  • $\begingroup$ The point or vertex A on the first triangle is equal to the vertex D on the second triangle. I was wondering if those points are equal, then the line segments created using those points are also equal. $\endgroup$ 2 days ago
  • $\begingroup$ You're saying that two points are equal (apparently dimensionally). But that can't be true, as points don't have dimensions. The points can only superimpose themselves on each other. $\endgroup$
    – Vanessa
    2 days ago
  • $\begingroup$ Write the full question. $\endgroup$
    – Vanessa
    2 days ago
  • $\begingroup$ I added the triangle picture, I hope it will help visualize the problem more clearly, as my wordings are kind of bad. $\endgroup$ 2 days ago

1 Answer 1

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In your image, you say that both triangles are equal in size. Then it follows that $AB=DE$.

Also, $SSS$ is side-side-side criterion, which means that the two triangles are congruent if all of their corresponding sides are equal. So, if you're using $SSS$ then you actually already had $AB=DE$. So you can't use this either.

The criterion that can be used here is $SAS$. If in your triangle it is given that $AC=DF$ and $BC=EF$ and $\angle ACB=\angle DFE$ then by corresponding parts of congruent triangles you can say that $AB=DE$.

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