Abcd is quadrilateral and o is a point inside it. prove that oa+ob+oc+od is less than ab+bc+cd+da.
Answers
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".
