Question

AB is a line segment.P and Q arevpointon 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

Hi malayali☺,

$answer$

Given P is equidistant from points A and B

PA=PB        .....(1)

and Q is equidistant from points A and B

QA=QB        .....(2)

In △PAQ and △PBQ

AP=BP  from (1)

AQ=BQ from (2)

PQ=PQ  (common)

So, △PAQ≅△PBQ  (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC  (common)

△PAC≅△PBC  (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT   ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180∘ (linear pair)

∠ACP+∠ACP=180∘ from (4)

2∠ACP=180°

∠ACP=180°/2=90°

Thus, AC=BC and ∠ACP=∠BCP=90°

∴,PQ is perpendicular bisector of AB.

Hence proved

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Note:-

• Try to attatch the figure of questions like this,so that the question can be answered with ease.

Hope it helps you better✨

Malayali pwoliyalle✌

Follow cheyth koode koodikko✔

Brainliest aayi mark cheyyanum marakkalle⚡

✨@athiraammu✨

Answer 2

Step-by-step explanation:

$in \: figure \: ab \: is \: a \: straight \: line \: on \:$

$which \: pb \: and \: ap \: sides \: are \: standing$

$in \: triangle \: aop \: and \: bop$

$op \: is \: a \: bisector \: of \: angle \: apb \: and \: ab \:$

$line$

$then \: angle \: apo = bpo$

$(op \: is \: the \: bisector \: of \: angle \: apb)$

$ao = ob \: (op \: is \: perpendicular \: and \:$

$the \: bisector \: of \: ab)$

$angle \: aop = bop = 90 \: degree$

$op \: perpendicular \: ab \: making \: right \: angle$

$op = op \: (common)$

$then \: triangle \: aop \: congruent \: to \:$

$triangle \: bop$

$hence \: proved \: that \: op \: is \: bisector \: and$

$perpendicular \: of \: ab$

$thanks$