To construct the perpendicular bisector of the line segment between two points requires two circles, each centered on an endpoint and passing through the other endpoint ( operation 2 ).
22.
:Alternatively, denoting the midpoint of AB by M, the angle AMP is a right angle because the perpendicular bisector of a chord goes through the centre of the circle; see chord ( geometry ).
23.
That the intersection of the plane with the cone is symmetric about the perpendicular bisector of the line through " F " 1 and " F " 2 may be counterintuitive, but this argument makes it clear.
24.
As shown above, if a circle passes through two given points "'P "'1 and "'P "'2, its center must lie somewhere on the perpendicular bisector line of the two points.
25.
Euler also points out that "'O "'can be found by intersecting the perpendicular bisector of "'Aa "'with the angle bisector of angle "'QAO "', a construction that might be easier in practice.
26.
As in the diagram above, form a circle through points A and B, tangent to edge BC of the triangle ( the center of this circle is at the point where the perpendicular bisector of AB meets the line through point B that is perpendicular to BC ).
27.
The point P must be equidistant from A and A'and also equidistant from B and B', so P must lie on the intersection of the perpendicular bisectors of AA'and BB'and as a corollary, the perpendicular bisector of CC'falls on the same point.
28.
Consider the triangle ABC . A unique circle also contains the point ABC and center O . Since O is equidistant from A and B, it must lie on the perpendicular bisector of AB . The same for BC and AC . So, the three perpendicular bisectors are concurrent at O.
29.
:: : : Be careful of the case when P lies on the ( extended ) line AB; in this case the perpendicular bisectors of AA'and BB'will be coincident, and you need the perpendicular bisector of CC'to pick out a particular point on this line.
30.
We notice that \ scriptstyle \ mathbf { k } and \ scriptstyle \ mathbf { k ^ \ prime } have the same magnitude, we can restate the Von Laue formulation as requiring that the tip of incident wave vector, \ scriptstyle \ mathbf { k }, must lie in the plane that is a perpendicular bisector of the reciprocal lattice vector, \ scriptstyle \ mathbf { K }.
