Question

Show that in a quadrilateral abcd ab+bc+cd+da ac+bd

Answer 1

Step-by-step explanation:

• Given

        ABCD is a quadrilateral.

        Side of quadrilateral are AB, BC, CD and DA.

        Diagonal of quadrilateral are AC and BD.

• Let

        Diagonal of quadrilateral are intersect at point O.

• We know that sum of two side of a triangle is greater than third side

• First consider triangle ΔOAB

        AB< OA + OB        ...1)

        Second consider triangle ΔOBC

        BC< OB + OC        ...2)

        Third consider triangle ΔOCD

        CD< OC + OD        ...3)

        Fourth consider triangle ΔODA

        DA< OD + OA        ...4)

• Adding equation 1) ,equation 2),equation 3) and equation 4). We get

        AB +BC + CD + DA < 2OA +2OB +2OC +2OD

        We can write above this

        AB +BC + CD + DA < 2(OA +OB +OC +OD)

       AB +BC + CD + DA < 2(OA +OC + OB + OD)     ...5)

• Where

       OA + OC =AC

       OB + OD = BD

• So equation 5) can be written as

       AB +BC + CD + DA < 2(AC + BD) This is proved.