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Problem 3. To bisect a given angle (Figure 67), or in other words, to construct the bisector of a given angle or to draw its axis of symmetry. Between the sides of the angle, draw an arc DE of arbitrary radius centered at the vertex B. Then, setting the compass to an arbitrary radius, greater however than half the distance between D and E (see Remark to Problem 1), describe two arcs centered at D and E so that they intersect at some point F. Drawing the line BF we obtain the bisector of the angle ABC.

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I cant't figure out why the arbitrary radius centered at D and E have to be greater than half the distance between D and E. Can someone point out as to why it is so?

The remark to Problem 1 is here:

Remark. For three segments to serve as sides of a triangle, it is necessary that the greatest one is smaller than the sum of the other two

Omicron
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    Hint: get a pencil and paper and try to make this construction work with a radius smaller than half that distance. – Ethan Bolker Apr 26 '17 at 00:06
  • @EthanBolker I know that the remark was the triangle inequality. But what is the relation between the triangle inequality and making the arbitrary radius greater than half the distance between D and E. – Omicron Apr 26 '17 at 00:28

3 Answers3

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Answer to the OP's question (in comments) about why the remark on the triangle inequality is a clue: it provides a formal proof of the observation that the two arcs don't intersect:

If the distance $EF = DF$ in the picture is less than half the distance $DE$ then the lengths of the edges of "triangle" $DEF$ violate the triangle inequality. So that triangle doesn't exist.

Ethan Bolker
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  • Correct if I'm wrong but I think I get it now. Imagine G to be the midpoint of DE. Then we want EF and DF to be greater than DG or EG. If we do so the EF + FG will be greater than the greater side DE and therefore there would be a triangle DEF because we have satisfied the triangle inequality. – Omicron Apr 26 '17 at 00:47
  • @Omicron Essentially, yes. I say "essentially" because I think I recall that the existence of a triangle with given sides that satisfy the triangle inequality is a little hazy in Euclid, and may be one of the points Hilbert worked to clarify. The nonexistence when it fails is clear since Euclid does prove the triangle inequality. – Ethan Bolker Apr 26 '17 at 00:54
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If the radius was smaller than half the distance from D to E then the two arcs that you draw wouldn't intersect so you wouldn't get a point F.

Brian Borchers
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  • I know that the remark was the triangle inequality. But what is the relation between the triangle inequality and making the arbitrary radius greater than half the distance between D and E. – Omicron Apr 26 '17 at 00:22
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You won't have a point "$F$". If the radius is too small you'll just have two circles centred at $D$ and $E$ which don't intersect so you can't identify a point exactly half way between the two lines.

The easiest way to think about it is to imagine two tiny circles centred at $D$ and $E$. "Half way" is just the limit where it stops working.

Harambe
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  • I get that but is there a reason as to why the book referred to the Remark below? – Omicron Apr 26 '17 at 00:09
  • That's the "triangle inequality". If you draw some pictures it will be obvious as well. If you have a triangle with sides $a$ and $b$ fixed but the angle free, the longest value for the third side is when the angle between them is $180^\circ$ right? (draw a picture to convince yourself). Then the length of the third edge is $a+b$ and not longer. – Harambe Apr 26 '17 at 00:12
  • I know that it is the triangle inequality I just don't know the relation between the triangle inequality and making the arbitrary radius greater than half the distance between D and E. – Omicron Apr 26 '17 at 00:18