Question

Sum of distances of any vertex of a quadrilateral ABCD from the other 3 is same for all the 4 vertices. Then, ABCD must be?

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Answer 1

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

To prove: quadrilateral ABCD is a rectangle

Solution:

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…

Answer 2

Answer:

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