last updated by  Pinaki Biswas 9 months ago
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    if the sum of the distance of the vertex of a quadrilateral from the other three is same for all the four vertices, then prove that the quadrilateral is a rectangle.

    swastik pramanik

    Let ABCD be the quadrilateral

    Given: BA + DA + CA = AB + CB + DB = AC + BC + DC = AD + BD + CD

    To prove: quadrilateral ABCD is a rectangle


    As we know: BA + DA + CA = AC + BC + DC

    We get: BA – BC = DC – DA — (i)

    Similarly from, AB + CB + DB = AD + BD + CD

    We get: CB – CD = AD – AB — (ii)


    By adding (i) & (ii) we get,

    BA – BC + CB – CD = DC – DA + AD – AB

    BA – CD = DC – AB

    2*AB = 2*CD

    AB = CD

    Similarly we can prove for: BC = AD

    We have have proved that quadrilateral ABCD is a PARALLELOGRAM…


    Now to prove that quadrilateral ABCD is indeed a rectangle. We have to prove that the the diagonals of the quadrilateral are equal… So, for quadrilateral ABCD we have to prove AC = BD


    As we know: BA + DA + CA = AB + CB + DB

    AC = BD (because, CB = DA)


    We have proved that the opposite sides and the diagonals are equal…

    Hence, quadrilateral ABCD is a RECTANGLE…

    swastik pramanik

    NOTE: In the question angle CXY should be angle CYX because if it is angle CXY then angle AXY and CXY would meet at the point X itself.

    The locus of the point P will be a straight line.

    Use the link below:

    click on the play button on the bottom left corner of the screen…

    Ashani Dasgupta

    Suppose ABCD is the quadrilateral. Then

    AB + AC + AD = DA + DB + DC

    This implies AB + AC = DB + DC —(i)

    Similarly BA + BC + BD = CA + CB + CD

    That is BA + BD = CA + CD (ii)

    Substracting (ii) from (i) we have AC – BD = DB – CA

    Hence AC = BD

    Similarly AB = DC.

    Hence ABCD is a parallelogram.

    Think! How can you extend this argument to show that this parallelogram is indeed a rectangle?

    Pinaki Biswas


    To Prove : Quadrilateral ABCD is a rectangle.

    Construction : None required.

    Proof : We know that AB+AD+AC=AB+BC+BD=BC+CD+AC=CD+AD+BD.

    Thus, AB+AD=BC+CD; (1)

    AB+BC=CD+AD. (2)

    From subtracting (2) from (1), we get:




    So, BC=AD.

    Thus, ABCD is a parallelogram… (The proof is not completed yet)

    Since: AB=BC,

    BC is common,


    Angles ABC=BCD.

    By proving all the angles equal by the method of congruency above,

    All angles must be right angles.

    Thus, ABCD is a rectangle. (Proved)


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