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In my geometry class last year I remember putting down the statement in a column proof "That all isosceles are always and only similar to other isosceles". I do not remember what I was trying to prove. But, I do remember that I was stressed and that was the only thing I could think of and made a guess thinking I would probably get the proof wrong on my test.

Funny enough though, I didn't get the proof wrong and I was wondering if anyone could show a proof as to why this would be true. I mean I makes sense but, I do not see any way to prove it. Could you please explain how this is true?

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Re-reading your question, I see two possible interpretations of your statement.

First (and my original answer), "If △ABC is isosceles and △ABC~△DEF, then △DEF is isosceles." Two triangles are similar if and only if the three angles of one are congruent to the three angles of the other. Since a triangle is isosceles if and only if two of its angles are congruent, if a triangle is similar to an isosceles triangle, then it will also have two congruent angles and must be isosceles.

Second, "If △ABC and △DEF are isosceles, then they are similar." This is not true. Suppose one triangle has angles with measures 20°, 20°, and 140° and another other triangle has angles with measures 85°, 85°, and 10°. Both triangles are isosceles (since within each triangle, there is a pair of congruent angles), but the triangles are not similar (because the angles of one are not congruent to the angles of the other).

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    I edited $m$y answer to, I think, better fit what you were asking.2019-01-25