Question
Show that in a quadrilateral $ABCD, AB + BC + CD + DA > AC + BD$
✓
Answer
Given $ABCD$ is a quadilateral,
Construction join diagonais $AC$ and $BD.$ To show $AB + BC + CD + DA > AC + BD$
In $\triangle\text{ABC}, AB + BC > AC [$Sum of two sides of a triangle is greater than the third side$] ...(i)$
In $\triangle\text{BCD}, BC + CD > BD [$Sum of two sides of a triangle is greater than the third side$] ...(ii)$
In $\triangle\text{CDA}, CD + DA > AC [$Sum of two sides of a triangle is greater than the third side$] ...(iii)$
In $\triangle\text{DAB}, DA + AB + > BD [$Sum of two sides of a triangle is greater than the third side$] ...(iv)$
On adding Eqs. $(i), (ii), (iii) (iv),$
We get $2(AB + BC + CD + DA) > 2 (AC + BD) $
$⇒ AB + BC + CD + DA > AC + BD$
Construction join diagonais $AC$ and $BD.$ To show $AB + BC + CD + DA > AC + BD$
In $\triangle\text{ABC}, AB + BC > AC [$Sum of two sides of a triangle is greater than the third side$] ...(i)$
In $\triangle\text{BCD}, BC + CD > BD [$Sum of two sides of a triangle is greater than the third side$] ...(ii)$
In $\triangle\text{CDA}, CD + DA > AC [$Sum of two sides of a triangle is greater than the third side$] ...(iii)$
In $\triangle\text{DAB}, DA + AB + > BD [$Sum of two sides of a triangle is greater than the third side$] ...(iv)$
On adding Eqs. $(i), (ii), (iii) (iv),$
We get $2(AB + BC + CD + DA) > 2 (AC + BD) $
$⇒ AB + BC + CD + DA > AC + BD$
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