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.
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Answer 1

GIVEN:- In quadrilateral ABCD, AC is the diagonal

TO PROVE:- (i) AB + BC + CD + DA > 2AC

(ii) AB + BC + CD > DA

(iii) AB + BC + CD + DA > AC + BD.

SOLUTION:- (i) Here, in ∆ABC,

AB + BC > AC (By exterior angle theorem)-------(1)

Similarly, in ∆CDA,

CD+ DA > AC (By exterior angle theorem)-------(2)

Adding Equation (1) & (2), we get

AB + BC + CD + DA > AC + AC

=> AB + BC + CD + DA >2AC (Proved)-------(3)

(ii) Here, in ∆ABC,

AB + BC > AC (By exterior angle theorem)

=> AB+BC+ CD> AC+ CD

=> AB + BC + CD > DA (Proved)

(iii) Here, in ∆ABD,

AB + AD > BD (By exterior angle theorem)------(4)

Similarly, in ∆CBD,

CB + CD > BD (By exterior angle theorem)-------(5)

Adding Equation (4) & (5), we get

AB + BC + CD + DA >2BD ------------(6)

Adding Equation (3) and (6), we get

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

=> AB + BC + CD + DA > AC + BD. (Proved)

OM NAMAH SHIVAYA

Debasish Biswal