Question

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

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

$\textsf{\blue{ Given }}$ : ABCD is a quadrilateral.

$\textsf{\blue{ To Verify }}$ : AB + BC + CD + DA > 2( AC + BD )

$\textsf{\blue{ Verification }}$ :

Since sum of two sides in greater than the third side.

OA + OD > AD___(1)

Since sum of two sides in greater than the third side.

OD + OC > CD ____(2)

Since sum of two sides in greater than the third side.

OB + OC > BC _____(3)

Since sum of two sides in greater than the third side.

OA + OB > AB ____(4)

On adding all 4 equations we get

OA + OD + OD + OC + OB + OC + OA + OB > AD + CD + BC + AB

2OA + 2OC + 2OB + 2OD > AB + BC + CD + DA

2( OA + OC + OB + OD ) > AB + BC + CD + DA

2 ( AC + BD ) $\bold{ > }$ AB + BC + CD + DA.

$\textsf{\green{ Hence verified }}$

$\textsf{\red{ Hope it helps }}$

$\textbf{ Have great future ahead! }$

Answer 2

$dear \: friend \: here \: is \: your \ \\ : answer \: ok$
IN A QUADILATERAL AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

AO + OB + OA + OD + OC + OB + OC > AB + AC + BC + CD

2(AC+BD)>AB+AC+BC+CD

HENCE YES THE STATEMENT IS TRUE