Question

Diagonals AD and BC of a quadrilateral ABCD intersect each other at O. Prove that () AO+ OC + CD + AD> AC+ OD (i) AB + OC + CD+AD<2(AC+ BC)
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Answer 1

Step-by-step explanation:

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)

HENCE, PROVED

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