Question

$ABCD$ is a convex quadrilateral with
$AB+BC\le AC+CD$. Prove that $AB<AC$.

Answer 1

Let us suppose that $AB\ge AC.$

Then $A$ lies on or to the left of the line $L$. ( the perpendicular bisector of $BC$ )

From the question; $AB+BC\le AC+CD$

$\Rightarrow AB=CD$ and $AD=BC$

[ Opposite sides are equal ]

Let us consider that the sides are of ${}^{\prime }{a}^{\prime }cm$. Then the diagonal length is $a\sqrt{2}cm.$

$\Rightarrow AB<AC$ i.e, $a<a\sqrt{2}.$

Hence, the answer is $AB<AC.$