In ΔABC, if AB = AC and BE, CF are the bisectors of ∠B and ∠C respectively. Prove that ΔEBC ≅ ΔFCB and BE = CF.

Since in ΔABC, AB = AC            ∠ABC = ∠ACB ......(i)
Since CF and BE are angle bisectors of ∠C and ∠B,
     we get ∠ABE = ∠EBC ...... (ii)
        and ∠ACF = ∠FCB ...... (iii) 
Now from eqns (i), (ii) and (iii), we get
1  ∠ABC =  1  ∠ACB
2               2
     󐺼 = ∠FCB ...... (iv)
Now in ΔFBC and ΔECB,
we have ∠FBC = ∠ECB   ...... (∠B = ∠C)
              BC = BC ...... (Common)
           󐿋 = ∠EBC ...... [From (iv)]
           ΔEBC ≅ ΔFCB
              BE = CF ...... (c.p.c.t.)