AMSCO'S Geometry. New York - Rye High School

200 Congruence Based on TrianglesConstruction 5Construct a Line Perpendicular to aGiven Line Through a Given Point on the Line.Given Point P on AB g.Construct PE g ' AB gEEACPDBACPDBACPDB1. With P as center andany convenient radius,draw arcs that intersectPA hat C and PB hat D.2. With C and D as centersand a radiusgreater than that usedin step 1, draw arcsintersecting at E.3. Draw EP g.ConclusionPE g ' AB gProofSince points C and D were constructed usingcongruent radii, CP = PD and P is equidistantto C and D. Similarly, since E was constructedusing congruent radii, CE ED, and E isequidistant to C and D. If two points are eachEequidistant from the endpoints of a line segment,then the points determine the perpendic-ABPular bisector of the line segment (Theorem 5.2).Therefore, PE gis the perpendicular bisector ofC DCD. Since CD is a subset of line AB g, PE g ' AB g.