Question

In the adjoining figure, ∠ABC = ∠ACB. M and N are points on the sides AC and AB respectively such that BN = CM. Prove that :

1. ΔNBC ≅ ΔMCB 
2. ΔPNB ≅ ΔPMC 
3. PB = РС.

Answer 1

Given:

In ΔABC, ∠ABC = ∠ACB.

Points M ∈ AC and N ∈ AB are such that BN = CM.

Let BM and CN meet at P.

To Prove:

1. ΔNBC ≅ ΔMCB 
2. ΔPNB ≅ ΔPMC 
3. PB = PC

Proof [Step-wise]:

i. ΔNBC ≅ ΔMCB

1. NB = CM   ...(Given)

2. BC = CB   ...(Common side)

3. ∠NBC = ∠ABC because N lies on AB and ∠MCB = ∠ACB because M lies on AC. 

Given ∠ABC = ∠ACB.

So ∠NBC = ∠MCB.

4. Therefore in triangles NBC and MCB we have side–angle–side (NB = CM, included angle ∠NBC = ∠MCB and BC common).

Hence ΔNBC ≅ ΔMCB.   ...(SAS)

ii. ΔPNB ≅ ΔPMC

Here, P is the intersection of BM and CN.

So PN lies on CN and PM lies on BM.

1. From (i), corresponding parts of congruent triangles give NC = MB and also corresponding angles equal in particular ∠CNB = ∠BMC and ∠MBN = ∠NCM.

2. ∠PNB = ∠CNB because PN is a part of CN; ∠PMC = ∠BMC because PM is a part of BM. 

By (i) ∠CNB = ∠BMC. 

Hence ∠PNB = ∠PMC.

3. ∠PBN = ∠MBN (PB lies on BM) and ∠PCM = ∠NCM (PC lies on CN). 

By (i) ∠MBN = ∠NCM.

Hence ∠PBN = ∠PCM.

4. BN = CM   ...(Given)

5. In triangles PNB and PMC we now have two angles equal (∠PNB = ∠PMC and ∠PBN = ∠PCM) and the included side between those two angles BN = CM.

Hence, by ASA, ΔPNB ≅ ΔPMC.

iii. PB = PC

1. From (ii), corresponding sides PB and PC are equal.

Therefore, PB = PC.

1. ΔNBC ≅ ΔMCB,
2. ΔPNB ≅ ΔPMC, 
3. PB = PC, as required.