In the given figure, O is the centre of the circle in which OD perpendicular to AC and OE perpendicular to BC and OD = OE. Show that triangle DBA is congruent to traingle EAB. Share with your friends Share 8 Manbar Singh answered this We have, CA and CB as the chords of the circle that is having the centre at O.Also OD⊥CA and OE⊥CB.Also , OD = OE Given⇒CA = CB Chords equidistant from the centre are equal in lengthWe know that ⊥ drawn from the centre to the chord bisects the chord.Since, OD⊥AC, thenAD = DC = 12CA ......1Since, OE⊥CB, thenBE = EC = 12CB .....2Now, CA = CB Proved above⇒ 12CA = 12CB ⇒AD = BENow, CA = CB⇒∠B = ∠A Angles opposite to equal sides are equalIn ∆DBA and ∆EABAD = BE Proved above∠A = ∠B Proved aboveAB = AB Common⇒∆DBA ≅ ∆EAB SAS 8 View Full Answer Amir answered this 30 -2