Question

AB is a line segment Pand Qare points on opposite sides of AB such that each of them is equidistant from the 1 point A and B so that the line PQ is perpendicular bisector of AB.

This is long question answer please .​

Answer 1

${\large{\bf{\underline{\underline{\purple \: {→Given :-}}}}}}$

• PA = PB AND QA = QB

${\large{\bf{\underline{\underline{\purple \: {→To \: Show:-}}}}}}$

• PQ ⊥ AB AND PQ BISECTS AB

NOW,

• Let us take (triangle) PAQ AND (Triangle) PBQ .

• In this triangles

• AP = BP. (GIVEN)
• AQ = BQ (GIVEN)
• PQ = PQ (COMMON)

So, (Triangle) PAQ ≡ (Triangle) PBQ (SSS Rule)

Therefore, <APQ =<BPQ (CPCT)

Now let us consider (Triangle) PAC = (Triangle) PBC

WE, HAVE

• AP = BP (GIVEN)

• <APC =<BPC [<ABQ =<BPQ Proved above]

• PC = PC (COMMON)

• So, (Triangle) PAC ≡ (Triangle) PBC [SAS]

• Therefore , AC = BC [CPCT] (1)

→ And, <ACP = <BCP [ CPCT]

Also, <ACP +<BCP = 180⁰ [LINEAR PAIR]

SO, 2 <ACP = 180⁰

OR

<ACP = 90⁰

HENCE PROVED