• # question_answer If ABCD is a quadrilateral whose diagonals AC and BD intersect at O, then A)  $\left( AB+BC+CD+DA \right)<\left( AC+BD \right)$ B)  $\left( AB+BC+CD+DA \right)>2\left( AC+BD \right)$ C)  $\left( AB+BC+CD+DA \right)>\left( AC+BD \right)$ D)  $AB+BC+CD+DA=2\,\left( AC+BD \right)$

In $\Delta \,\,ABC,\,\,\Delta \,\,ACD,\,\,\Delta \,\,BCD$and $\Delta \,\,ABD$ AB + BC > AC CD + DA > AC BC + CD >BD DA + AB > BD Adding above inequalities 2 (AB + BC + CD +DA) > 2 (AC+ BD) $\Rightarrow$(AB + BC +CD DA) > (AC + BD)