Question

in the given figure ABC is a line segment p and q are points on opposite sides of a b such that each of them is equidistant from the points a and b show that the line PQ is the perpendicular bisector of AB​

Answer 1

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

• ab is a line segment p and q are point on opposite sides of ab such that each of them is equidistant from the point a and b. show that line pq is perpendicular bisector of ab

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A line segment AB and two points P and Q such that PA = PB and QA = QB.

${\orange{\sf\underline{To\:Prove}}}$

Let AB and PQ intersect at C.Then,we have to prove that Ac = BC and ∠ACP = 90°

${\pink{\sf\underline{Proof}}}$

In ∆PAQ and ∆PBQ,we have

PA = PB (given)

QA = QB (given)

PQ = PQ (commom)

∴ ∆PAQ ≅ ∆PQB (by SSS-criteria).

∴ ∠APQ = ∠BPQ ⠀⠀⠀⠀⠀⠀...(i) ⠀(c.p.c.t.).

Now,in ∆PAC and ∆PBC , we have

PA = PB (given)

∠APC = ∠BPC [∴ ∠APQ = ∠BPQ in (i) ]

PC = PC (commom)

∴ ∆PAC ≅ ∆PBC (by SAS-criteria).

∴ AC = BC ⠀⠀⠀⠀⠀⠀⠀.....(ii)⠀(c.p.c.t.).

And,∠ACP = ∠BCP⠀⠀⠀⠀⠀....(iii)⠀(c.p.c.t.).

But,∠ACP + ∠BCP = 180°⠀(linear pair).

∴ 2∠ACP = 180° ⠀[ using (iii) ].

So,∠ACP = 90°.

Hence,PQ is the perpendicular bisector of AB

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