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# 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.

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

**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

**☛ Related Questions:**

- Show that in a right-angled triangle, the hypotenuse is the longest side.
- In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
- In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
- In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

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