Geometry: An Interactive Journey to Mastery

Lesson 12: Equidistance—A Focus on Distance

x The set of points equidistant from a pair of intersecting lines is precisely the set of points on an angle

bisector of those lines.

x The three angle bisectors of a triangle coincide at a single point. This point is equidistant from all three

sides of the triangle and, therefore, lies at the center of a circle inside the triangle tangent to all three of

its sides.

Summary

Identifying the set of all points of equal distances from a given set of objects provides a powerful tool for

proving results in geometry. In this lesson, we examine equidistance between points and between lines and

prove the existence of special circles for triangles. We also solve some standard schoolbook geometry problems

in sophisticated ways.

Example 1

/HWA ͼ௘௘ͽDQGB ͼ௘௘ͽ

D௘ͽ )LQGWKHHTXDWLRQRIWKHSHUSHQGLFXODUELVHFWRURIAB.

E௘ͽ )LQGDSRLQWP on this line with xFRRUGLQDWHHTXDOWR

F௘ͽ 8VHWKHGLVWDQFHIRUPXODWRYHULI\WKDWP is indeed equidistant from A and B.

Solution

D௘ͽ 7KHSHUSHQGLFXODUELVHFWRUSDVVHVWKURXJKWKHPLGSRLQWͼ௘௘ͽDQGLWVVORSHLVWKHQHJDWLYHUHFLSURFDO

of slope AB ^84 2.

Its equation is thus yx^252 1 , or yx 25.^12

E௘ͽ 3XWx = 10 to see that y 55,^12 giving y  2259 2.

We have P §· ̈ ̧©¹10,^92.

F௘ͽ PA   (^7492) ̈ ̧§·©¹ 2443 2 9 205 and PB   39.^2 ©¹ ̈ ̧§·^132442169205
These are indeed equal.