Consider \(\displaystyle\triangle{B}{A}{D},\triangle{C}{A}{E}\)

As \(\displaystyle\overline{{{A}{D}}}\perp\overline{{{D}{B}}},\angle{B}{D}{A}={90}^{{\circ}}\)

As \(\displaystyle\overline{{{A}{E}}}\perp\overline{{{E}{C}}},\angle{C}{E}{A}={90}^{{\circ}}\)

So,

\(\displaystyle\angle{C}{E}{A}\stackrel{\sim}{=}\angle{B}{D}{A}\) (i) (right angles are congruent to each other)

Also,

\(\displaystyle\angle{B}{A}{D}\stackrel{\sim}{=}\angle{C}{A}{E}\) (ii) (common angles in \(\displaystyle\triangle{B}{A}{D},\triangle{C}{A}{E}\))

As \(\displaystyle\angle{A}{B}{C}\stackrel{\sim}{=}\angle{A}{C}{B}\),

\(\displaystyle\overline{{{A}{B}}}\stackrel{\sim}{=}\overline{{{A}{C}}}\) (iii)

(In a triangle, sides opposite to equal angles are equal)

From (i), (ii) and (iii),

\(\displaystyle\triangle{B}{A}{D}\stackrel{\sim}{=}\triangle{C}{A}{E}\) (by AAS congruence criteria)