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Construction
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11.2
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To construct the perpendicular bisector of a given line segment.

Given a line segment AB, we want to construct its perpendicular bisector.

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Steps of Construction:
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- Taking A and B as centres and radius more than 1/2 AB, draw arcs on both sides of line segment AB (to intersect each other).
- Let These arcs intersect each other at P and Q. Join PQ.
- Let PQ intersect AB at the Point M.

Then line PMQ is the required Perpendicular bisector of AB.

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Justification
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We have to prove PQ is perpendicular bisector,
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i.e.
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AM = BM and ∠ PMA = ∠ PMB = 90°
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Join A,B to P and A, B to Q

In Δ PAQ and Δ PBQ,

AP = BP

AQ = BQ

PQ = PQ

∴ ∆ PAQ ≅ ∆PBQ

So, ∠ APM = ∠ BPM

Now ,

In Δ PAM and Δ PBM,

AP = BP

PM = PM

∠APM = ∠BPM

∴ ∆ PMA ≅ ∆PMB

So, AM = BM & ∠ PMA = ∠ PMB

Also,

∠PMA + ∠PMB = 180°

∠PMA + ∠PMA = 180°

2∠PMA = 180°

∠ PMA = 90°

Thus, BM & ∠ PMA = ∠ PMB = 90°

∴ PQ is the perpendicular bisector of AB