Point of intersection of a triangle on a Cartesian plane?

Prove algebraically that the three perpendicular bisectors of the sides of the triangle ABC are concurrent (all pass through one point). Prove that the point of intersection of the three perpendicular bisectors of the sides of triangle ABC is the point P that is being sought i.e. that it is equidistant from A, B and C.

Can someone please explain the working out?

Thank you

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  • 5 years ago
    Best Answer

    Let PD & PE be the perpendicular bisectors of BC & CA respectively.

    Drop a perpendicular PF on AB.

    To prove AF = FB.

    BD = DC.

    PD common.

    Included angle = 90

    So ∆PDB ≡ ∆PDC

    So PD = PC.

    Similarly ∆PEC ≡ ∆PEA

    So PC = PA

    We get, PC = PB = PA

    So P is equidistant from A, B & C.

    in ∆APF & ∆BPF,

    PA = PB, <FAP = <PBF

    By construction <AFP = <BFP = 90

    So ∆AFP ≡ ∆BFP

    So AF = FB Hence Proved.

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