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Q. ABCD is a quadrilateral. Prove that (AB + BC + CD + DA) > (AC + BD)

 

Dear Chesna,

For a shorter solution, you can follow this:

Let ABCD be a quadrilateral with AC and BD be its diagonals.

In ΔABC,

AB + BC > AC  (Sum of any two sides of a triangle is greater than the third side)

In ΔACD,

CD + AD > AC

Adding both the equation, we get

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

⇒ (AB + BC + CD + AD) > 2AC

Similarly, it can be proved hat (AB + BC + CD + AD) > 2BD by taking the triangle ABD and BCD.

Now adding these two results, we get

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

⇒ (AB + BC + CD + DA) > (AC + BD ).. Hence Proved!

Hope it helps!


Regards!