Question
$ABCD$ is a quadrilateral. Is $AB + BC + CD + DA < 2(AC + BD) ?$
✓
Answer
In \triangle $OAB,$
$OA + OB > AB. . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (i)$
In $\triangle OBC,$
$OB + OC > BC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
In $\triangle OCD,$
$OC + OD > CD. . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iii)$
In $\triangle OAD,$
$OA + OD > AD . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (iv)$
Adding $(1), (2), (3)$ and $(4)$
$2 (OA + OB + OC + OD) > AB + BC + CD + DA$
$ \therefore AB + BC + CD + DA < 2(OA + OB + OC + OD)$
$ \therefore AB + BC + CD + DA < 2(OA + OC + OB + OD)$
$ \therefore AB + BC + CD + DA < 2(AC + BD)$
Hence proved.
$OA + OB > AB. . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (i)$
In $\triangle OBC,$
$OB + OC > BC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
In $\triangle OCD,$
$OC + OD > CD. . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iii)$
In $\triangle OAD,$
$OA + OD > AD . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (iv)$
Adding $(1), (2), (3)$ and $(4)$
$2 (OA + OB + OC + OD) > AB + BC + CD + DA$
$ \therefore AB + BC + CD + DA < 2(OA + OB + OC + OD)$
$ \therefore AB + BC + CD + DA < 2(OA + OC + OB + OD)$
$ \therefore AB + BC + CD + DA < 2(AC + BD)$
Hence proved.
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