[Mathukumilli Sumanth]

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

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...

[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:

https://ggbm.at/ZqV9RHv8

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

[Dr. 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]

Given : AB+AD+AC=AB+BC+BD=BC+CD+AC=BC+AD+BD

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:

AB-CD=CD-AB.

So,AB=CD.

Similarly,BC=AD.

So, BC=AD.

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

Since: AB=BC,

BC is common,

AC=BD,

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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