Question

If ABCD is a quadrilateral then show AB+BC+CD+DA <2 (AC+BD)

Answer 1

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 ) > AB + BC + CD + DA.

$\textsf{\green{ Hence verified }}$ 

$\textsf{\red{ Hope it helps }}$ 

$\textbf{ Have great future ahead! }$ 

Note : I have answered same question few days before,

https://brainly.in/question/6609925

( The link contains my answer )

Answer 2

in triangle ABC
AB+BC>CA........(1)
in triangle ABD
AB+AD>BD.........(2)
in triangle CBD
BC+CD>BD........(3)
in triangle ADC
AC+DC>A.........(4)
END