Question

06 ABCD is a quadrillateral Prove that (AB+BC +CD+DA), (AC+BD) ​

Answer 1

Answer:

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