Draw a line segment `6.4 cm ` long and draw its perpendicular bisector. Measure the length of each part.

Explanation

STEPS OF CONSTRUCTION
(i) Draw a ling segment `AB=6.4` cm.
(ii) With `A` as centre and a radius equal to more than half of `AB`, draw two arcs, one above `AB` and the other below `AB.`
(iii) With `B` as centre and the same radius, draw two arcs, cutting the previously drawn arcs at points `C` and `D` respectively.
(iv) Join `CD` is the required perpendicular bisector of AB at the point O.
On measuring, we find that
`OA=3.2` cm and `OB=3.2` cm
Also, `angleAOC=angle BOC=90^(@)`
Justification :
Join AC, AD, BC and BD.
In `Delta CAD` and `Delta CBD`, we have
`AC=BC` (arcs of equal radii)
`AD=BD` (arcs of equal radii)
`CD=CD` (common)
`:. Delta CAD cong Delta CBD` (SSS-criteria)
`:. angle ACO = angle BCO` (c.p.c.t.).
Now, in `Delta AOC` and `Delta BOC`, we have
`AC=BC` (arcs of equal radii)
`angleACO=angle BCO` (proved above)
`CO=CO` (common)
`:. Delta AOC cong Delta BOC` (SAS-criteria).
Hence, `AO=BO` and `angleAOC=angleBOC`.
But, `angleAOC+angleBOC=180^(@)` (linear pair axiom)
`:. angle AOC=angle BOC=90^(@)`.
Hence, COD is the perpendicular disector of `angleAOB`.