Question

in a quadrilateral ABCD in which diagonal AC and BD intersect at O show that a b + BC + CD + D A is less than 2 ( AC + BD )

Answer 1

Step-by-step explanation:

ABCD is a quadrilateral and AC & BD are the diagonals.

As we know,

The sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,

In Δ AOB , AB < OA + OB  ……….(i)

In Δ BOC , BC < OB + OC  ……….(ii)

In Δ COD, CD < OC + OD  ……….(iii)

In Δ AOD , DA < OD + OA  ……….(iv)

⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD

⇒ AB + BC + CD + DA < 2[( AO + OC) + (DO + OB)]

⇒ AB + BC + CD + DA < 2(AC +  BD)

Hence, proved.

Hope the answer was helpful for ya!

Answer 2

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)