If O is a point within a quadrilateral ABCD show that OA+OB+OC+OD is greater than AC+BD
monalisha132:
hyy
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Answered by
11
Given : ABCD is a quadrilateral. O is a point inside the quadrilateral ABCD.
To prove : OA + OB + OC + OD > AC + BD
Construction : Join OA, OB, OC and OD. Also, join AC and BD
Proof : By triangle in equality the sum of any two sides of a triangle is greater than the third side.
In ΔBOD,
OB + OD > BD …........(1)
Similarly
In ΔAOC,
OA + OC > AC ….........(2)
Adding (1) and (2), we obtain
OB + OD + OA + OC > BD + AC
∴ OA + OB + OC + OD > AC + BD
To prove : OA + OB + OC + OD > AC + BD
Construction : Join OA, OB, OC and OD. Also, join AC and BD
Proof : By triangle in equality the sum of any two sides of a triangle is greater than the third side.
In ΔBOD,
OB + OD > BD …........(1)
Similarly
In ΔAOC,
OA + OC > AC ….........(2)
Adding (1) and (2), we obtain
OB + OD + OA + OC > BD + AC
∴ OA + OB + OC + OD > AC + BD
but it's wrong
bcuz BOD is a line not a triangle
yes sry actually it was just misunderstanding so sry
it is really a triangle. I made wrong construction so I was confused
Answered by
7
Let ABCD BE A QUADRILATERAL WHOSE DIAGONALS R AC AND BD ABD O ANY POINT WITHIN QUADRILATERAL .
JOIN O WITH A,B,C,D
IN ACO
OA+OC > A.C. ...... ( 1 )
( IN triangle sum of 2 SIDESGREATER than 3rd sides )
similarly in BOD
OB+OD > BD......... (2)
ADDING BOTH EQUATION WE GET
OA+OC +OB+OD> AC+ BD .
HOPE IT HELP YOU
tnks 2
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