The following problem seems obvious:
If $\triangle ABC$ and $\triangle DEF$ are such that $|AB|=|DE|$ and $|BC|=|EF|$ but $\angle ABC > \angle DEF$ then $|AC|>|DF|$.
But I can't to write a formal proof!! Any suggestion, thanks in advance!!
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Valent
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3To aid in web-searching, this is called the "Hinge Theorem". – Blue Aug 31 '14 at 22:21
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1This is helpful!! Thanks. – Valent Aug 31 '14 at 22:24
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Do you know the Law of Cosines? – JimmyK4542 Aug 31 '14 at 22:50
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Yes! But I was looking for a classical approach!! – Valent Sep 01 '14 at 04:36
