We have

To show that, MN ∥ BC.

Given that, ∠B = ∠C and BM = CN.

So, AB = AC [sides opposite to equal angles (∠B = ∠C) are equal]

Subtract BM from both sides, we get

AB – BM = AC – BM

⇒ AB – BM = AC – CN

⇒ AM = AN

⇒ ∠AMN = ∠ANM

[angles opposite to equal sides (AM = AN) are equal] …(i)

We know in ∆ABC,

∠A + ∠B + ∠C = 180° [∵ sum of angles of a triangle is 180°] …(ii)

And in ∆AMN,

∠A + ∠AMN + ∠ANM = 180° [∵ sum of angles of a triangle is 180°] …(iii)

Comparing equations (ii) and (iii), we get

∠A + ∠B + ∠C = ∠A + ∠AMN + ∠ANM

⇒ ∠B + ∠C = ∠AMN + ∠ANM

⇒ 2∠B = 2∠AMN [∵ from equation (i), and also ∠B = ∠C]

⇒ ∠B = ∠AMN

Thus, MN ∥ BC since the corresponding angles, ∠AMN = ∠B.