Question

ABCD is a Quadrilateral. Is AB + BC + CD + DA > 2 (AC + BD ) ?

Answer 1

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ABCD is a quadrilateral and AC, and BD are the diagonals. Sum of the two sides of a triangle is greater than the third side. So, considering the triangle ABC, BCD, CAD and BAD, we get AB + BC > AC CD + AD > AC AB + AD > BD BC + CD > BD Adding all the above equations, 2(AB + BC + CA + AD) > 2(AC + BD) ⇒ 2(AB + BC + CA + AD) > 2(AC + BD) ⇒ (AB + BC + CA + AD) > (AC + BD) ⇒ (AC + BD) < (AB + BC + CA + AD)
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Answer 2

Since, the sum of length of any two sides in a triangle should be greater than the Length of third side.


Therefore, In ΔAOB,
$\: \: \: \: \bold{AB < OA + OB}$ ..........(i)

In ΔBOC,
$\: \: \: \: \bold{BC < OB + OC}$ ..........(ii)

In ΔCOD,
$\: \: \: \: \bold{CD < OC + OD}$ ..........(iii)

In ΔAOD,
$\: \: \: \: \bold{DA < OD + OA}$ ..........(iv)


$\text{\underline{Adding equations (i), (ii), (iii) and (iv) we get:}}$


$\textbf{AB + BC + CD + DA < OA + OB + OB +OC + OC + OD + OD + OA}$


$⇒\textbf{AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD}$


$⇒\textbf{AB + BC + CD + DA < 2[(AO + OC) + (DO +OB)]}$


$\bold{\red{\fbox{AB + BC + CD + DA < 2(AC + BD)}}}$


Hence, it is proved ✔✅


[ Based on NCERT solutions ]