Question

Show 2(AC + BD) > AB + BC +CD + DA.

Answer 1

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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