1.

(b) ABCD is a quadrilateral in which the diagonals AC and BD intersect at 0. Prove thatAB+BC + CD + AD < 2(AC + BD).hc

Answer»

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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 getAB + BC > AC ..........(1)CD + AD > AC .........(2)AB + AD > BD .........(3)BC + CD > BD .........(4)

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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