Plzz help me

loca95

New member
Joined
Feb 14, 2012
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15
ABCD est un parallélogramme
MONTRE QUE : AB+AD=<AC+BD

I need the solution so much plzz help
 
This is saying that the sum of the diagonals is larger than the sum of two of the connecting sides. What are you at liberty to use in proving this? Please show what you've tried and where you are stuck.
 
This is saying that the sum of the diagonals is larger than the sum of two of the connecting sides. What are you at liberty to use in proving this? Please show what you've tried and where you are stuck.

Honestly a have no idea even about how to start it :(
Geometry is always killing me
 
Do you know what the definition of a parallelogram is? Geometrically, what properties does it have? Do you know the triangle inequality?
 
Draw a parallelogram, label the vertices as given. Look at the diagonals, and look at the sides. Try to find a way to get the inequality.

For example, if the parallelogram happened to be a square of all side length 1, then AB + AD = \(\displaystyle 2\). AC + BD = \(\displaystyle 2\sqrt{2}\).

Try it when it is a rectangle of side lengths 1,2,1,2.
 
Draw a parallelogram, label the vertices as given. Look at the diagonals, and look at the sides. Try to find a way to get the inequality.

For example, if the parallelogram happened to be a square of all side length 1, then AB + AD = \(\displaystyle 2\). AC + BD = \(\displaystyle 2\sqrt{2}\).

Try it when it is a rectangle of side lengths 1,2,1,2.

Sorry But I have no idea :|
 
no just because i need it for an exam tomorrow
im really sorry :(

I'm sorry but you don't seem to know anything about this question. My hints will lead you to an answer if you give some effort. This is a "help" site...
 
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