Constructions
59. The Perfect Split · Constructing the perpendicular bisector
CD bisects AB at M and is perpendicular to AB.
The perpendicular bisector of segment AB is constructed by drawing two arcs of equal radius (greater than ½AB) from A and B. The two arc intersections C and D define a line CD that bisects AB at its midpoint M AND is perpendicular to AB.
What this lesson covers
Try to break it
Drag A and B! As long as AB is less than twice the radius, the intersection points C and D define the perpendicular bisector. Try making AB too long and see what happens.
How you build it
Construct the perpendicular bisector of a segment.
- Draw the line segment AB.
- Set your compass to a radius greater than half of AB. Draw an arc from A.
- Without changing the radius, draw an arc from B.
- Join the intersection points C and D with a straight line.
The proof, step by step
Prove that CD bisects AB at right angles.
- Draw CD. In ΔACD and ΔBCD, AC=BC and AD=BD (equal radii), CD is common.
- So ΔACD ≅ ΔBCD by SSS. This means ∠ACD = ∠BCD.
- Now in ΔAMC and ΔBMC, AC=BC, ∠ACM=∠BCM, and CM is common.
- So ΔAMC ≅ ΔBMC by SAS. Thus AM=BM and ∠AMC=∠BMC.
- Since ∠AMC + ∠BMC = 180°, each must be 90°. Hence CD ⊥ AB.
Worked example
While constructing the perpendicular bisector of a line segment AB, why is it necessary to take the radius of the arcs more than half of AB?
If the radius is less than or equal to half of AB, the arcs will not intersect (or will touch at one point), making it impossible to define the bisector line.
- To make the arcs intersect at two distinct points. — correct
- To ensure the compass does not slip.
- To make the construction faster.
- It does not matter what the radius is.