In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:

(i) △DBC ≅ △ECB

(ii) ∠DCB = ∠EBC

(iii) OB = OC, where O is the point of intersection of BE and CD.

(i) From figure,

BD = CE (Given)

BC = BC (Common)

∠DBC = ∠ECB (as AB = AC).

∴ △DBC ≅ △ECB by SAS axiom.

Hence, proved △DBC ≅ △ECB.

(ii) We know that corresponding parts of congruent triangle are equal.

∴ ∠DCB = ∠EBC.

Hence, proved that ∠DCB = ∠EBC.

(iii) As △DBC ≅ △ECB,

∠BDO = ∠CEO (By c.p.c.t.)

∠DOB = ∠EOC (Vertically opposite angles)

BD = CE (Given)

Hence, △BOD ≅ △EOC by ASA axiom.

We know that corresponding parts of congruent triangles are equal.

∴ OB = OC.

Hence, proved that OB = OC.