Question

P
M
5. In the adjoining figure, AB is a line segment. P and Q are
points on the 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.
A plz answer correctly otherwise will be reported!!!!!

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Answer 1

Step-by-step explanation:

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=

2

180

∘

=90

∘

Thus, AC=BC and ∠ACP=∠BCP=90

∘

∴,PQ is perpendicular bisector of AB.

Hence proved.

Answer 2

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 A PAQ and A PBQ

AP=BP from (1)

AQ=BQ from (2)

PQ=PQ (common)

So, A PAQ: A PBQ (SSS congruence)

Hence ZAPQ=ZBPQ by CPCT

In A PAC and A PBC

AP=BP from (1)

LAPC=ZBPC from (3)

PC=PC (common)

A PAC= A PBC (SAS congruence)

::AC=BC by CPCT