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A B is a line segment and line l is it...

`A B` is a line segment and line `l` is its perpendicular bisector. If a point `P` lies on `l ,` show that `P` is equidistant from `A` and `Bdot`

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Given: AC=BC
`/_PCA=/_PCB=90^@`
To prove: `PA=PB`
Proof: In `trianglePCA` and `trianglePCB`
AC=BC [Given]
`/_PCA=/_PCB=90^@` [Given]
`PC=PC` [common]
`triangleAPC~=triangleABC` [`SAS`]
`=>PA=PB`
Hence proved.
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