BC is a line segment. A and D are points on opposite sides of BC such that each of them is
equidistant from the points B and C. Show that the line AD is the perpendicular bisector of
BC.
Answers
Answer:
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=
2
180
∘
=90
∘
Thus, AC=BC and ∠ACP=∠BCP=90
∘
∴,PQ is perpendicular bisector of AB.
Hence proved.
Answered By
toppr
611 Views
How satisfied are you with the answer?
This will help us to improve better
answr
Get Instant Solutions, 24x7
No Signup required
girl
More Questions by difficulty
EASY
MEDIUM
HARD
star
Answer:
a. nzndbslirbsofaoDjksoqjs a qok