# Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

**Solution:**

Given, ABCD is a quadrilateral

We have to show that AB + BC + CD + DA < 2 (BD + AC)

Join the diagonals AC and BD of the quadrilateral.

Consider the triangle OAB,

We know that the sum of two sides of a triangle is greater than the third side.

OA + OB > AB ------------ (1)

Consider the triangle OBC,

We know that the sum of two sides of a triangle is greater than the third side.

OB + OC > BC ------------ (2)

Consider the triangle OCD,

OC + OD > CD ------------ (3)

In triangle ODA,

OD + OA > DA -------------- (4)

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

OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA

2(OA + OB + OC + OD) > AB + BC + CD + DA

From the figure,

AC = OA + OC

BD = OB + OD

2[(OA + OC) + (OB + OD)] > AB + BC + CD + DA

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

The above expression can be rewritten as

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

Therefore, it is proven that AB + BC + CD + DA < 2(BD + AC)

**✦ Try This:** In the figure, it is given that AE = AD and BD = CE. Prove that △AEB is congruent △ADC.

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 7

**NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 11**

## Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

**Summary:**

It is shown that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

**☛ Related Questions:**

visual curriculum