# AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.

**Solution:**

Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD

To prove: ∠A > ∠C and ∠B > ∠D

Let's join vertex A to C as shown below.

In the above ΔABC,

AB < BC (given AB is the smallest side of quadrilateral ABCD)

∠ACB < ∠BAC (Angle opposite to the smaller side is smaller) ... (1)

In ΔADC,

AD < CD (given CD is the largest side of quadrilateral ABCD)

∠ACD < ∠CAD (Angle opposite to the smaller side is smaller) ... (2)

On adding Equations (1) and (2), we obtain

∠ACB + ∠ACD < ∠BAC + ∠CAD

∠C < ∠A

This means, ∠A > ∠C

Hence Proved.

Let's now join BD.

In ΔABD,

AB < AD (Given AB is the smallest side of quadrilateral ABCD)

∠ADB < ∠ABD (Angle opposite to the smaller side is smaller) ...(3)

In ΔBDC,

BC < CD (Given CD is the largest side of quadrilateral ABCD)

∠BDC < ∠CBD (Angle opposite to the smaller side is smaller) ... (4)

On adding equations (3) and (4), we obtain

∠ADB + ∠BDC < ∠ABD + ∠CBD

∠D < ∠B

This means, ∠B > ∠D

Hence, proved.

**Video Solution:**

## AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.

### NCERT Maths Solutions Class 9 - Chapter 7 Exercise 7.4 Question 4:

**Summary:**

If AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD, then we have proved that ∠A >∠C and ∠B >∠D.