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I found the following theorem in a lecture notes without proof: Let $A, B, C, D, E$ and $F$ be points on the plane such that $\angle ABC$ and $\angle DEF$ are either both acute or they are both obtuse. If $AB$ is perpendicular to $DE$ and $BC$ is perpendicular to $EF$ then $\angle ABC=\angle DEF$. I was able to see the acute angle case but why do those both angles can both be obtuse? And how to prove the obtuse case?

geometrystudent
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IF you know spanish, this page will be usefull:

https://es.wikipedia.org/wiki/Arco_capaz

After you draw a sketch for the obtuse angle case, consider the complementary angles and those are equal because they share an arc of a circumference. Then the complementary angles are equal.

PenasRaul
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  • Sorry but I don't understand Spanish. This theorem was given in the notes before the theorem that the angles which share the same arc are equal. – geometrystudent Sep 29 '14 at 18:22
  • Then you have to do as I said, consider the acute angles complementary to the obtuse angles, these angles also are in the conditions stated, so they must be of equal amplitude. – PenasRaul Sep 29 '14 at 19:21