In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that $A B + B C + C D + D A > 2 A C$ .

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Solution

Given $A B C D$ is a quadrilateral and $A C$ is one of its diagonal.
So:-
(i) In $Δ A B C; A B + B C > A C$ ……… $(1)$
(ii) In $Δ A C D; C D + D A > A C$ ………. $(2)$
By adding $(1)$ & $(2)$
$A B + B C + C D + D A \geq 2 A C$ .

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Similar questions

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.

A B C D

A B + B C + C D + D A < 2 (A C + B D)

A B C D

A B + B C + C D + D A > 2 A C

Let ABCD be a quadrilateral in which the diagonals intersect at O perpendicularly. Prove that AB + BC + CD + DA > AC + BD.

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