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.
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:
$\endgroup$ 2Remark. 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
3 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
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