Question

In the given figure, AB is a line segment P and Q are points on opposite sides of AB 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

Answer:

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Answer 2

$\huge\underline{\underline{\texttt{{Question:}}}}$

• In the given figure, AB is a line segment P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is the perpendicular bisector of AB

$\huge\underline{\underline{\texttt{{Given:}}}}$

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

$\huge\underline{\underline{\texttt{{To\:prove:}}}}$

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

$\huge\underline{\underline{\texttt{{Proof:}}}}$

$in \: \triangle \:PAQ \: and \: \triangle \: PBQ \: we \: have$

• PA = PB (given)
• QA = QB (given)
• PQ = PQ (common)

$\therefore \: \triangle \: PAQ \: \cong \: \triangle \: PBQ \: (by \: sss - criteria).$

$\therefore \: \angle \: APQ = \angle \: BPQ \: \: \: \: \: ....(i) \: \: (c.p.c.t.).$

Now, in ∆PAC and ∆PBC, we have

• PA = PB (given)
• ∠APC = ∠BPC [∴ ∠APQ = ∠BPQ in (i)]
• PC = PC (common)

$\therefore \: \angle \: PAC \: \cong \: \angle \: PBC \: \: (by \: SAS - criteria)$

$\therefore \: AC = BC \: \: \: \: \: \: \: \: ....(ii) \: (c.c.p.t.)$

$and \: \angle \: ACP \: = \: \angle \: BCP \: \: \: \: \: ....(iii) \: \: (c.c.p.t.)$

But,

• ∠ACP = ∠BCP = 180° (linear pair)

$\therefore \: 2\angle \: ACP = 180 \degree \: \: [using \: (iii)].$

So,

• ∠ACP = 90°.

$\huge\underline{\underline{\texttt{{Hence:}}}}$

• PQ is the perpendicular bisector of AB.