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  • Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

Answers (1)

Given : ABCD is a quadrilateral

To prove : AB + BC + CD + DA + < 2 (BD + AC)

Proof: Since we know that the sum of lengths of any two sides in a triangle should be greater than the third side.

\therefore In \triangleAOB, AB < OA + OB

In \triangleBOC, BC < OB + OC

In \triangleCOD, CD < OC + OD

In \triangleAOD, DA < OD + OA

Adding all of the above,

\Rightarrow AB + BC + CD + AD < 2OA + 2OB + 2OC + 2OD

\Rightarrow AB + BC + CD + AD < 2((AO + OC) + (DO + OB))

\Rightarrow AB + BC + CD + AD < 2(AC + BD) 

(\because AC = AO + OC, BD = DO + OB)

Hence proved.

Posted by

infoexpert24

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