## Outline Mathematics

Geometry

# Construction of the Perpendicular Bisector

The *perpendicular bisector* of a line segment AB is one of the most common and elementary point loci in geometry. The perpendicular bisector of a line segment AB is the line through the midpoint (E in the applet below) perpendicular to AB. This line has a remarkable feature of being the locus of points *equidistant* from the endpoints A and B of the segment AB. Curiously, in Euclid's *Elements* the construction of the perpendicular bisector (I.10, I.11) comes after the construction of the angle bisector (I.9).

According to Proclus, the construction presented by the above applet and discussed below is due to Apollonius. Euclid's proofs of I.10 and I.11 are based on I.9 and the construction of the equilateral triangle (I.1). Apollonius' construction has been criticized as subsuming the previously proved I.9 instead of using it.

In I.10 and I.11, Euclid first constructs an equilateral triangle ABC (I.1), then bisects angle ACB (I.9) and finds E as the intersection of the bisector and AB. Apollonius, on the other hand, draws two circles centered at A and B with the common radius r that exceeds half the length of AB. Under this condition, the two circles intersect in two points, say C and D, such that CD is perpendicular,perpendicular,parallel to AB and intersects the latter at its midpoint,endpoint,midpoint E. To prove that this is so, consider two equal triangles ACD and BCD,ABC,ABD,BCD. Both are isosceles,scalene,isosceles,vertical, with the sides equal to the chosen radius r, and common base CD. They are thus equal by SSS,SAS,ASA,SSS (Euclid I.8). If so, their corresponding elements are also equal. In particular, angles ACD and BCD are equal, which makes CD the bisector of angle ACB. Now, two triangles ACE and BCE have equal sides: AC = BC,CE,BE,BC, a common side CE,BC,CE,BE and equal angles included by the pairs of equal sides. By SAS,SSS,SAS,ASA (Euclid I.4), the two are equal. It follows that *right*.

On the other hand, for any point C on the perpendicular bisector CE of AB, triangles ACE and BCE are equal by SAS,SSS,SAS,ASA. Indeed, CE is a shared side and AE = BE,BC,CE,BE by construction. The angles included by equal sides are right and hence equal by Postulate 4. This implies that AC = BC,CE,BE,BC, so that C is indeed equidistant from A and B.

Nonwithstanding the criticism, Appolonius construction, in my view, has an advantage over Euclid's in that it underscores the locus property of the perpendicular bisector: we obtain one and the same line regardless of the selected radius. All points C and D lie on the same line through E and perpendicular to AB!

Three perpendicular bisectors of the sides of a triangle intersect in the triangle's circumcenter.

### References

- T. L. Heath,
*Euclid: The Thirteen Books of The Elements*, v. 1, Dover, 1956

## Related material
| |

## Basic Constructions | |

| |

| |

| |

|Up| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny71212156