Verified answer

Correct Statment:-

ABCD is a quadrilateral.O is the point inside the quadrilateral.Prove that AB+BC+CD+DA < 2(OA+OB+OC+OD)

Given :-

ABCD is a quadrilateral.

O is the point inside the quadrilateral.

Required To Prove :-

OA + OB + OC + OD < AB + BC + CD + DA

Construction :-

Join OA , OB , OC and OD.

Join AC and BD

Proof :-

O is the inside of the quadrilateral ABCD,

We know that

" The sum of any two sides of a triangle is greater than the third side".

In ∆ AOB , OA+OB > AB ------(1)

In ∆ BOC , OB + OC > BC -----(2)

In ∆ COD , OC + OD > CD -----(3)

In ∆ DOA , OD + OA > DA -------(4)

On adding all above equations then

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

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

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

Therefore, AB+BC+CD+DA < 2(OA+OB+OC+OD)

It cane be further written as

OA + OC = AC and OB + OD = BD

Therefore,

(AB+BC+CD+DA) < (AC+BD)

Hence, Proved.

Used Inequality :-

♦ Inequality of a triangle : " The sum of any two sides of a triangle is greater than the third side".