5. To construct the perpendicular bisector of a line segment

In the construction of the perpendicular bisector of a line segment, we are

Usually, one makes all the arcs with the same radii. It is simpler that way. If that is the case, then the figure is a rhombus, and we know that the diagonals of a rhombus are perpendicular bisectors of each other. But the top and bottom triangles do not have to be congruent. They just have to be isosceles. If, for instance, after you have made the arcs intersect at *C* it may happen that arcs with the same radius would intersect off the paper below the line, and in that case you could adjust the compass so that the arc would intersect at a point *D* which was on the paper.

We have access to a theorem which will make this easier to prove. We are given that *C* and *D* are equidistant from *A* and *B*, and we know that a point is on the perpendicular bisector of a line segment if and only if it is equidistant from its endpoints. Since *C* and *D* are on the perpendicular bisector of *AB, CD* is the perpendicular bisector of *AB.*