Question

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

Answer 1

Answer:

Step-by-step explanation:

This question is based on the triangle inequality theorem that the sum of lengths of two sides of a triangle is always greater than the third side.

Now visually identify that the quadrilateral ABCD. It is divided by diagonals AC and BD into four triangles.

Now, take each triangle separately and apply the above property and then add L.H.S and R.H.S of the equation formed.

In triangle ABC,

AB + BC > AC ……………(1)

In triangle ADC,

AD + CD > AC. (2)

In triangle ADB,

AD + AB > DB --(3)

In triangle DCB,

DC+ CB > DB..(4)

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

AB + BC + AD + CD + AD + AB + DC+ CB > AC + AC + DB + DB

AB + AB + BC + BC + CD + CD + AD + AD > 2AC + 2DB  

2AB + 2BC + 2CD + 2AD > 2AC + 2DB

AB + BC + CD + AD > AC + DB

Hence, AB + BC + CD + DA > AC + BD is true

Useful Tip:

Whenever you encounter problems of this kind, it is best to think of the property based on the sum of lengths of any two sides of a triangle is always greater than the third side.

Answer 2

Given : ABCD is a quadrilateral and AC and BD are diagonals

To prove : ( AB + BC + CD + DA ) > (AC + BD)

Solution : ABCD is a quadrilateral and AC ,and BD are diagonals.

Sum of the two sides of a triangle is greater tghan the third side .

So, considering the triangle ABC ,BCD ,CAD and BAD.

We get,

AB + BC > AC

CD + AD >AC

AB + AD > BD

BC + CD > BD

Adding all the above equations

2 ( AB + BC + CA + AD ) > 2 ( AC + BD )

= 2 ( Ab + BC + CA + AD ) > 2 ( AC + BD)

= ( AB + BC + CA + AD ) > ( AC + BD )

= Hence proved