Is it true that if ABCD is quadrilateral, then AB + BC + CD + DA < 2 (AC + BD)?

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True

false

answer is A.

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Detailed Solution

We have been given a quadrilateral ABCD.
Question Image

We know that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In triangle AOB, we have,

A B < O A + O B … … … (1)

In triangle COD, we have,

C D < O C + O D … … ... (2)

In triangle AOD, we have,

D A < O D + O A … … … (3)

In triangle COB, we have,

B C < O C + O B … .. … . (4)

Adding equations (1), (2), (3), and (4), we get,

A B + B C + C D + D A < O A + O B + O C + O D + O D + O A + O C + O B ⇒ A B + B C + C D + D A < 2 O A + 2 O B + 2 O C + 2 O D ⇒ A B + B C + C D + D A < 2 (O A + O C) + 2 (O B + O D) (5)

From figure,

O A + O C = A C and O B + O D = B D .

Therefore, from equation (5), we have,

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

Therefore, the given statement is true.
Hence, option (1) is correct. As a result, PM stands for altitude.

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