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# In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

**Solution:**

Given: ∠B < ∠A and ∠C < ∠D

To prove: AD < BC

We can use the fact that in any triangle, the side opposite to the larger (greater) angle is longer.

In ΔAOB, ∠B < ∠A (given)

AO < OB (The side opposite to the smaller angle is smaller) ... (1)

In ΔCOD, ∠C < ∠D

OD < OC (The side opposite to the smaller angle is smaller) ... (2)

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

AO + OD < BO + OC

AD < BC

Hence, proved

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

**Video Solution:**

## In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC

NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.4 Question 3

**Summary:**

If in the given figure, ∠B < ∠A and ∠C < ∠D, we have seen that AD < BC.

**☛ 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.
- 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.
- In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

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