question_answer 1)

Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of \[{{l}_{2}}\] and \[{{l}_{1}}\] Let them meet at P. Is PA = PB?

Answer:

                Here, we will use the following steps of construction: Step I Draw any angle XOY with vertex O. Step II Take a point A on OX and a point B on OY, such that OA = OB.                   Step III Now, with O as centre, using compasses, draw an arc radius more than half of the length of \[{{l}_{1}}\] Step IV With the same radius and with A as centre, draw another arcs using compasses. Which cut the previous arc at C and D respectively. Step V Join \[{{l}_{2}}\] It cuts \[S\to \] at M. Therefore, \[Y\to \] is the perpendicular bisector of \[M\to \] Step VI Now, with O as centre and radius equal to more than half of the length of OB, draw arcs. Step VII With same radius and with B as centre, draw another arcs which cut the previous arcs at E and F. Step VIII Join \[E\to \] It cuts \[T\to \] at N. Therefore, EF is the perpendicular of \[R\to \]. Also, both perpendicular bisectors meet at A. On measuring, we get PA = PB.