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

- 56

Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.

Therefore,

In

In

In

In

Adding the equations (1) , (2) , (3) and (4) we get,

PQ + QR + PS + SR + RQ + PS + PQ > PS + PS + SQ + SQ

=>( PQ + PQ ) + ( QR + QR ) + ( PS + PS ) + ( SR + SR ) > 2 PR + 2 SQ

=> 2 PQ + 2 QR + 2 PS + 2 SR > 2 PR + 2 SQ

=> 2 ( PQ + QR + PS + SR) > 2 ( PR + SQ )

=> PQ + QR + PS + SR > PR + SQ

=> PQ + QR + RS + SP > PR + SQ

Hence, it is true.

- 3

IN TRIANGLE ABC

AB+BC IS GREATER THAN AC

IN TRIANGLE ADC

CD+DA IS GREATER THAN AC

THUS AB+BC+CD+DA IS GREATER THAN 2AC

IN TRIANGLE ADB

AD+AB IS GREATER THAN BD

IN TRIANGLE BDC

CD+BC IS GREATER THAN BD

THUS AB+BC+CD+DA IS GREATER THAN 2BD

THUS AB+BC+CD+DA+AB+BC+CD+DA IS GREATER THAN 2(AC + BD)

= 2(AB+BC+CD+DA) IS GREATER THAN 2(AC +BD)

THUS AB+BC+CD+DA IS GREATER THAN AB+BD

HENCE IT PROVED

- 0

_{SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER TAN THE THIRD SIDE}_{IN TRIANGLE ABC}_{AB+BC IS GREATER THAN AC}_{IN TRIANGLE ADC}_{CD+DA IS GREATER THAN AC}_{THUS AB+BC+CD+DA IS GREATER THAN 2AC}_{IN TRIANGLE ADB}_{AD+AB IS GREATER THAN BD}_{IN TRIANGLE BDC}_{CD+BC IS GREATER THAN BD}_{THUS AB+BC+CD+DA IS GREATER THAN 2BD}_{THUS AB+BC+CD+DA+AB+BC+CD+DA IS GREATER THAN 2(AC + BD)}_{= 2(AB+BC+CD+DA) IS GREATER THAN 2(AC +BD)}_{THUS AB+BC+CD+DA IS GREATER THAN AB+BD}_{HENCE IT PROVED}

- 1

- 1