Question -
Answer -
Yes, Given Undefined control sequence \squareABCD in which AC and BD are its diagonals.
In ∆ABC, we have
AB + BC > AC …(i)
[Sum of any two sides is greater than the third side]
In ∆BDC, we have
BC + CD > BD …(ii)
[Sum of any two sides is greater than the third side]
In ∆ADC, we have
CD + DA > AC …(iii)
[Sum of any two sides is greater than the third side]
In ∆DAB, we have
DA + AB > BD …(iv)
[Sum of any two sides is greater than the third side]
Adding eq. (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2CD + 2DA > 2AC + 2BD or AB + BC + CD + DA > AC + BD [Dividing both sides by 2]
Hence, proved.