Question

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

Answer 1

Solution :

Let us prove a lemma , the sum of two sides of a triangle is always greater than the third side.

Consider any triangle ABC .

Extend the side length AB to D to form another triangle , DBC

Now , utilising these properties , we can always show that the sum of two sides is any triangle is greater than the third side .

In the given figure , consider ∆ABC

AB + BC > AC ....... (1)

Now consider ∆ ADC

AD + DC > AC ........ (2)

Now consider ∆ ABD

AB + AD > BD ........ (3)

And finally in ∆ BCD

BC + CD > BD ...... (4)

Adding all these 4 equations :

AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD

=> 2 { AB + BC + CD + DA } > 2 { AC + BD }

2 being a constant gets cancelled .

Hence , we obtain that , AB + BC + CD + DA > AC + BD

Or that the sum of sides of any quadrilateral is always greater than the sum of its diagonals .

This holds true for all cases .

Thus proved

_________________________________________