Question -
Answer -
Given : In quadrilateral ABCD, AC and BD are its diagonals,
┬а
To prove : AB + BC + CD + DA > AC + BD
Proof: In тИЖABC,
AB + BC > AC тАж(i)
(Sum of any two sides of a triangle is greater than its third side)
Similarly, in тИЖADC,
DA + CD > AC тАж(ii)
In тИЖABD,
AB + DA > BD тАж(iii)
In тИЖBCD,
BC + CD > BD тАж(iv)
Adding (i), (ii), (iii) and (iv)
2(AB + BC + CD + DA) > 2AC + 2BD
тЗТ 2(AB + BC + CD + DA) > 2(AC + BD)
тИ┤ AB + BC + CD + DA > AC + BD