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Question
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2(BD + AC)
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Solution
Given: ABCD is a quadrilateral.
To show: AB + BC + CD + DA < 2(BD + AC)
Construction: Join diagonals AC and BD.
Proof: In ΔOAB, OA + OB > AB ...(i) [Sum of two sides of a triangle is greater than the third side]
In ΔOBC, OB + OC > BC ...(ii) [Sum of two sides of a triangle is greater than the third side]
In ΔOCD, OC + OD > CD ...(iii) [Sum of two sides of a triangle is greater than the third side]
In ΔODA, OD + OA > DA ...(iv) [Sum of two sides of a triangle is greater than the third side]
On adding equations (i), (ii), (iii) and (iv), we get
2[(OA + OB + OC + OD] > AB + BC + CD + DA
⇒ 2[(OA + OC) + (OB + OD)] > AB + BC + CD + DA
⇒ 2(AC + BD) > AB + BC + CD + DA ...[∵ OA + OC = AC and OB + OD = BD]
⇒ AB + BC + CD + DA < 2(BD + AC)
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