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Question
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
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Solution
In ΔAOB, we have
OA + OB > AB ...(i)
In ΔBOC, we have
OB + OC > BC ...(ii)
In ΔCOD, we have
OC + OD > CD ...(iii)
In ΔAOD, we have
OA + OD > AD ....(iv)
Adding (i), (ii), (iii) and (iv), we get
2 (OA + OB + OC + OD) > AB + BC + CD + AD
⇒ 2 [(OA + OC) + (OB + OD)] > AB + BC + CD + AD
⇒ 2 (AC + BD) > AB + BC + CD + AD
[∵ OA + OC = AC and OB + OD = BD]
⇒ AB + BC + CD + AD < 2 (AC + BD).
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