O ia a point inside a quadrilateral abcd....prove that oa+ob+oc+od>ac+bd
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Let in quad. abcd,o is a point inside it
To prove: oa+ob+oc+od> ac+bd
Construction:Diagonals ac and bd are joined
Proof:In triangle aoc,
oa+oc>ac(since, sum of any two sides of a triangle is greater than the third side). (1)
Similarly,in triangle boc,
ob+od>bd(since,sum of any two sides of a triangle>third side (2)
Adding equation (1)and (2),
oa+oc+ob+od>ac+bd
Therefore,oa+ob+oc+od>ac+bd
To prove: oa+ob+oc+od> ac+bd
Construction:Diagonals ac and bd are joined
Proof:In triangle aoc,
oa+oc>ac(since, sum of any two sides of a triangle is greater than the third side). (1)
Similarly,in triangle boc,
ob+od>bd(since,sum of any two sides of a triangle>third side (2)
Adding equation (1)and (2),
oa+oc+ob+od>ac+bd
Therefore,oa+ob+oc+od>ac+bd
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