TO CONSTRUCT THE BISECTOR OF A LINE SEGMENT

We draw a line segment of length 7.8 cm draw the perpendicular bisector of this line segment.

AB = 7.8 cm.

STEPS:

• Draw the line segment AB = 7.8 cm.
•  With point A as centre and a suitable radius, more than half the length of AB, draw arcs on both the sides of AB.
• With point B as centre and with the same radius draw arcs on both the sides of AB. Let these arc cut at points P & Q as shown on in the figure.
• Draw a line through the points P and Q. The line so obtained is the required perpendicular bisector of given line segment AB.

Line PQ is perpendicular bisector of AB.

(A) PQ bisects AB i.e., OA = OB.

(B) PQ is perpendicular to AB i.e., ∠PAO = ∠POB = 90o.

    Proof : In ΔAPQ and ΔBPQ :

AP = BP                [By construction]

AQ = BQ                [By construction]

PQ = PQ                [Common]

⇒    ΔAPQ = ∠BPQ                [By SSS]

⇒    ∠APQ = ∠BPQ                [By cpctc]

Now, in Δ APO & ΔBPO

AP = BP                [By construction]

OP = OP                [Common side]

∠APO = ∠BPO                [Proved above]

⇒    ΔAPOΔBPO            [By SAS]

And, ∠POA = ∠POB

=180º/2= 90^o            [∠POA + ∠POB = 180^o]

⇒    PQ is perpendicular bisector of AB.