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.

Question

​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

Manbar Singh · Accepted Answer

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

​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.

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.