26
Property of the perpendicular bisector of a segment
Textbook: pp. 66–67
GoalKnow the perpendicular bisector of a segment and its property.
New words
perpendicular bisector · o‘rta perpendikulyarequidistant · teng uzoqlikdamidpoint · kesma o‘rtasiproperty · xossa
Explanation
A line through the midpoint of a segment and perpendicular to it is called the perpendicular bisector of the segment. Theorem: every point of the perpendicular bisector is equidistant from the ends of the segment. Proof: let O be the midpoint of AB and C a point on the perpendicular bisector. In △ACO and △BCO, CO is common, AO = BO and ∠AOC = ∠BOC = 90°; by SAS △ACO = △BCO, so AC = BC. This property helps to find equal segments in problems.
Worked examples
If C lies on the perpendicular bisector of AB and AC = 9 cm, then BC = 9 cm.
In △ABC the perpendicular bisector of BC meets AC at E; BE = 5, AC = 12, so EC = 5 and AE = 12 − 5 = 7.
Class activity
“Fold the bisector”: fold a paper strip so the endpoints meet and draw along the crease — this is the perpendicular bisector.
Practice
1
C lies on the perpendicular bisector of AB, BC = 14 cm. Find AC (cm).
14
2
In △ABC the perpendicular bisector of BC meets AC at E, BE = 8, AC = 20. Find AE.
12
3
What angle does the perpendicular bisector make with the segment?
90°
4
Why is a point of the perpendicular bisector equidistant from the ends?
The two triangles are equal by SAS, so AC = BC.