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 #21270
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…April 23, 2018 at 3:55 pm #21278
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…April 23, 2018 at 4:15 pm #21293
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?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)
From subtracting (2) from (1), we get:
Thus, ABCD is a parallelogram… (The proof is not completed yet)
BC is common,
By proving all the angles equal by the method of congruency above,
All angles must be right angles.
Thus, ABCD is a rectangle. (Proved)
You must be logged in to reply to this topic.