Diagonal AC and BD of quadrilateral ABCD intersects each other at point O.

Prove that AB + BC + CD + AD < 2 (AC + BD).

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Solution

In a quadrilateral ABCD, with diagonals intersecting at O. AO+OB>AB(INEQUALITY)

OD+OA>AD

OD+OC>DC

OB+OC>BC

ADDING

AO + OB + OA + OD + OC + OB + OC > AB + AC + BC + CD

(AC+BD+AC+BD)>AB+BC+CD+AD

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

HENCE YES THE STATEMENT IS TRUE

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A B C D

¯ ¯¯¯¯¯¯ ¯ A B = ¯ ¯¯¯¯¯¯¯ ¯ A D

¯ ¯¯¯¯¯¯ ¯ B C = ¯ ¯¯¯¯¯¯¯ ¯ D C

¯ ¯¯¯¯¯¯ ¯ A C

¯ ¯¯¯¯¯¯¯ ¯ B D

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

A B C D

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

A B C D

A B ∥ C D

A C

B D

O

\frac{O A}{O C} = \frac{O D}{O B}

A B^{2} + C D^{2} = B C^{2} + D A^{2}

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