# Are diagonals of parallelogram greater than its sides?

I was doing a question on rectangle and thought about the diagonals of it .

And just remembered about parallelogram and tried to prove or disprove that both the diagonals of parallelogram are greater than its sides but wasn't able to do so...

I have proved it for one diagonal but the second one is difficult for me to do

I let that parallelogram has one acute angle and one obtuse angle and diagonal opposing to obtuse angle makes a triangle in which it has to be greatest because it is opposite to greatest angle

For diagonal opposing acute angle, I tried cosine rule but it didn't help.

Please help to prove or disprove that both of diagonals of parallelogram are greater than its sides......!!!!

• If you squeeze a parallelogram it's clear that one diagonal can be as short as you like. – user4894 Mar 28 '17 at 4:13
• The smaller diagonal does not have to be greater than either side. Just take any parallelogram and "flatten" it until it degenerates into a segment - the sides stay the same, but the smaller diagonal tends to $0\,$. – dxiv Mar 28 '17 at 4:13
• Just take a rhombus and almost flatten it. The answer should then be visually obvious. – quasi Mar 28 '17 at 4:13

For simplicity let the parallelogram be a rhombus, all sides equal to $s$. And an acute angle of $\alpha$. Then the shorter diagonal is $s2\sin (\frac {\alpha}2)$.
So we need an an angle where $\sin (\frac {\alpha}2) <1/2$. That happens if $\frac {\alpha}2 < 30$ or $\alpha < 60$.