Question
$ABCD$ is a quadrilateral. Is $AB + BC + CD + DA > AC + BD?$
✓
Answer
In $\triangle ABC,$
$AB + BC > AC . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (i)$
In $\triangle ACD,$
$CD + DA >AC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
Adding $(i)$ and $(ii).$
$AB + BC + CD + DA > 2AC. . . . (iii)$
In $\triangle ABD$
$AB + DA > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iv)$
In $\triangle BCD$
$BC + CD > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (v)$
Adding $(iv)$ and $(v),$
$AB + BC + CD + DA > 2ABD . . . . (vi)$
Adding $(iii)$ and $(vi),$
$2 [AB + BC + CD + DA] > 2(AC + BD)$
$ \therefore AB + BC + CD + DA > AC + BD.$
$AB + BC > AC . . . $[Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (i)$
In $\triangle ACD,$
$CD + DA >AC . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (ii)$
Adding $(i)$ and $(ii).$
$AB + BC + CD + DA > 2AC. . . . (iii)$
In $\triangle ABD$
$AB + DA > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side]$ . . . . (iv)$
In $\triangle BCD$
$BC + CD > BD . . .$ [Sum of the lengths of any two sides of a triangle is greater than the length of the third side] $. . . . (v)$
Adding $(iv)$ and $(v),$
$AB + BC + CD + DA > 2ABD . . . . (vi)$
Adding $(iii)$ and $(vi),$
$2 [AB + BC + CD + DA] > 2(AC + BD)$
$ \therefore AB + BC + CD + DA > AC + BD.$
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