∠BAE=∠CAE {Angles inscribed in the same arc}

∴arc BE=arc CE

∴arc BE=arc CE∴ chord BE≅ chord CE ___(1)

(1)In △BDE and △CDE,

(1)In △BDE and △CDE, BE=CE {from (1)}

BD=CD {ED is perpendicular bisector of BC}

DE=DE {common side}

∴△BDE≅△CDE {SSS test of congruence}

⇒∠BDE=∠CDE {C.P.C.T}

Also, ∠BDE+∠CDE=180∘ {angles in linear pair}

∴ ∠BDE = ∠CDE = 90∘

90∘

90∘ Therefore, DE is the right bisector of BC.