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
(||) AB + BC + CD >DA
(iii) AB + BC + CD + DA > AC + BD​

Answer 1

Answer:

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Step-by-step explanation:

(i) Consider △ ABC

We know that AB + BC > AC ……

(1) Consider △ ACD We know that AD + CD > AC ……. (2)

By adding both the equations we get AB + BC + AD + CD > AC + AC So we get AB + BC + AD + CD > 2AC ……..

(3) Therefore, it is proved that AB + BC + AD + CD > 2AC. (ii) Consider △ ABC We know that AB + BC > AC Add CD both sides of the equation AB + BC + CD > AC + CD …… (4) Consider △ ACD We know that AC + CD > DA ……. (5) By substituting (5) in (4) we get AB + BC + CD > DA …….. (6) (iii) Consider △ ABD and △ BDC We know that AB + DA > BD …….

(7) So we get BC + CD > BD …….

(8) By adding (7) and (8)

we get AB + DA + BC + CD > BD + BD

On further calculation

AB + DA + BC + CD > 2BD …….

(9) By adding equations (9) and (3)

we get

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

So we get 2 (AB + BC + CD + DA) > 2 (BD + AC)

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

Therefore, it is proved that AB + BC + CD + DA > BD + AC.

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