Math Olympiad, I.S.I. Entrance and College Mathematics › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Geometry › prove that the quardilateral is a rectangle

This topic contains 5 replies, has 4 voices, and was last updated by Pinaki Biswas 5 months, 3 weeks ago.

- AuthorPosts
- April 23, 2018 at 1:47 pm #21254
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.

April 23, 2018 at 3:01 pm #21270Let 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…

April 23, 2018 at 3:55 pm #21278NOTE: 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…

- This reply was modified 5 months, 3 weeks ago by Ashani Dasgupta.
- This reply was modified 5 months, 3 weeks ago by Ashani Dasgupta.

April 23, 2018 at 4:15 pm #21293Suppose 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?

April 28, 2018 at 3:05 pm #21380**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)** - AuthorPosts

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