Question

ABCD is a quadrilateral. Prove that : AB+BC+CD+DA<2(AC+BD)

Answer 1

hey friend your answer is ###
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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 any two sides of a triangle is greater than the third side

therefor , in triangle ABC , we have
A B +BC > AC. eq 1
In triangle BCD , we have eq 2
BC +CD> BD
In triangle CDA ,we have. eq 3
DA+CD> AC
In triangle DAB ,We have. eq 4
DA + AB > BD
Adding eq 1,2,3 and 4 ,we get
2 AB+2BC + 2CD + 2 DA > 2 AC + 2 BD
AB + BC + AD + DA > AC + BD