Question

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

Answer 1

Answer:

Yes.

Step-by-step explanation:

Let us consider ABCD is quadrilateral and P is the point where the diagonals are intersect.

We 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 1]

Then, consider the ΔPBC

Here, PB + PC < BC … [equation 2]

Consider the ΔPCD

Here, PC + PD < CD … [equation 3]

Consider the ΔPDA

Here, PD + PA < DA … [equation 4]

By adding equation [1], [2], [3] and [4] 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.