Question

9. In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD.
Please solve this question​

Answer 1

Knowing the triangle is important.

Solution: The sum of two sides is always greater than one side.

Question (i)

As ΔABC and ΔACD forms, we can use this to prove the inequality.

→AB+BC>AC and CD+DA>AC

→Find their sum, so the proof is complete.

Question (ii)

As ΔABC forms, we get an inequality.

→AB+BC>AC

→AB+BC+CD>AC+CD [Adding CD on both sides]

→AB+BC+CD>AC+CD>DA [A triangle ΔACD forms]

→∴AB+BC+CD>DA Hence proved.

Question (iii)

As ΔABC and ΔBCD forms, we have the two.

→AB+BC>AC and BC+CD>BD

→Find their sum, so the proof is complete.

Answer 2

Given: ABCD is a quadrilateral and AC is one of its diagonal.

(i) We know that the sum of any two sides of a triangle is greater than the third side.

In ∆ ABC, AB + BC > AC----(1)

In ∆ ACD, CD + DA > AC----(2)

Adding inequalities (1) and (2), we get:

AB + BC + CD + DA > 2AC

(ii) In ∆ ABC, we have :

AB + BC > AC----(1)

We also know that the length of each side of a triangle is greater than the positive difference of the length

of the other two sides.

In ∆ ACD, we have:

AC > |DA − CD|----(2)

From (1) and (2), we have:

AB + BC > |DA − CD|

⇒ AB + BC + CD > DA

(iii) In ∆ ABC, AB + BC > AC

In ∆ ACD, CD + DA > AC

In ∆ BCD, BC + CD > BD

In ∆ ABD, DA + AB > BD

Adding these inequalities, we get:

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

⇒ (AB + BC + CD + DA) > (AC + BD)