Diagonal AC and BD of a quadrilateral ABCD intersect each other at O. Prove that
1) AB+BC+CD+DA>AC+BD
2)AB+BC+CD+DA>2(AC+BD)

Dear student,

1)



Your 2nd part appears to be incorrect recheck your question and get back to us for meaningful help.


Regards

  • -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)
  • 0
What are you looking for?