Question

ABCD is a quadrilateral.
Is AB + BC + CD + DA >AC + BD?
ABCD is quadrilateral. Is
AB + BC + CD + DA<2 (AC + BD)?​

Answer 1

2)know that,

The sum of the length of any two sides is always greater than the third side.

Now consider the ΔPAB,

Here, PA + PB > AB … [equation i]

Then, consider the ΔPBC

Here, PB + PC > BC … [equation ii]

Consider the ΔPCD

Here, PC + PD > CD … [equation iii]

Consider the ΔPDA

Here, PD + PA > DA … [equation iv]

By adding equation [i], [ii], [iii] and [iv] we get,

PA + PB + PB + PC + PC + PD + PD + PA > AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD > AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD > AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) > AB + BC + CD + DA

From the figure we have, AC = PA + PC and BD = PB + PD

Then,

2AC + 2BD > AB + BC + CD + DA

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

Hence, the given expression is true..

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)

HENCE, PROVED

NOTE - YOUR 1ST QUESTION IS 2ND OF MY ANSWER.

Answer 2

Question:

ABCD is a quadrilateral.

Is AB + BC + CD + DA >AC + BD?

Ans- In 1st attachment.

ABCD is quadrilateral. Is

AB + BC + CD + DA<2 (AC + BD)?

Ans- In 2nd attachment.