question_answer

ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

Answer:

Yes, given ABCD be a quadrilateral and its diagonal AC and BD intersect each other at O. We know that, the sum of lengths of any two sides of a triangle is greater than the length of third side. In ΔAOB, OA + OB > AB ?.(i) Similarly, In ΔBOC, OB + OC > BC ??.. (ii) In ΔCOD           OC + OD > CD ??. (iii) And In ΔDOA,    OD + OA > DA ?? (iv) On adding Eqs. (i), (ii), (iii) and (iv), we get             \[\left( OA+OB \right)+\left( OB+OC \right)+\left( OC+OD \right)+(OD+\] OA) > AB + BC + CD + DA \[2\left( OA+OB+OC+OD \right)>AB+BC+CD+DA\] \[AB+BC+CD+DA<2\left( OA+OB+OC+OD \right)\] \[AB+BC+CD+DA<2\left[ \left( OA+OC \right)+\left( OB+OD \right) \right]\] AB + BC + CD + DA < 2(AC + BD)   \[\left[ \text{ }AC=OA+OC\text{ }and\text{ }BD=OB+OD \right]\]