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# Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

**Solution:**

Given, ABCD is a quadrilateral

We have to show that AB + BC + CD + DA > AC + BD

Join the diagonals AC and BD of the quadrilateral.

Considering triangle ABC,

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

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

Considering triangle BCD,

BC + CD > BD ---------------- (2)

Considering triangle CDA,

CD + DA > AC ---------------- (3)

Considering triangle DAB,

DA + AB > BD ----------------- (4)

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

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

On rearranging,

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

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

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

**✦ Try This:** In figure, AD=BD and ∠C=∠E. Prove that BC=AE.

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

**NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 12**

## Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

**Summary:**

It is shown that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD

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