Question

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

Answer 1

ABCD is a quadrilateral then prove thatAB + BC + CD + DA < 2(AC + BD)?ABCD is a quadrilateral and AC, andBD are the diagonals. Sum of the two sides of a triangle is greater than the third side.

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 ΔABC,
$\: \: \: \: \textbf{AB + BC > AC}$ .......... (i)

In ΔADC,
$\: \: \: \: \textbf{AD + DC > AC}$ .......... (ii)

In ΔDCB,
$\: \: \: \: \textbf{DC + CB > DB}$ .......... (iii)

In ΔADB,
$\: \: \: \: \textbf{AD + AB > DB}$ .......... (iv)


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


$\textbf{AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB}$

$⇒\textbf{(AB+AB) + (BC+BC) + (AD+AD) + (DC+DC) > 2AC + 2DB}$

$⇒\textbf{2AB + 2BC + 2AD + 2DC > 2(AC+DB)}$

$⇒\textbf{2(AB + BC + AD + DC) > 2(AC+DB)}$

$⇒\textbf{AB + BC + AD + DC > AC + DB}$


$\bold{\red{\fbox{AB + BC + CD + DA > AC + DB}}}$


Hence, it is true ✔✅


[ Based on NCERT solutions ]