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