(i) Based on the △ ABD and △ ACD

From △ ABC we know that AB and AC are equal sides of isosceles triangle

So we get

AB = AC

From △ DBC we know that DB and DC are equal sides of isosceles triangle

So we get

DB = DC

We also know that AD is common i.e. AD = AD

Therefore, by SSS congruence criterion we get

△ ABD ≅ △ ACD

(ii) We know that △ ABD ≅ △ ACD

We get ∠ BAD = ∠ CAD (c. p. c. t)

It can be written as

∠ BAE = ∠ CAE ……… (1)

Considering △ ABE and △ ACE

We know that AB and AC are the equal sides of isosceles △ ABC

AB = AC

So by using equation (1) we get

∠ BAE = ∠ CAE

We know that AE is common i.e. AE = AE

Therefore, by SAS congruence criterion we get

△ ABE ≅ △ ACE

(iii) We know that △ ABD ≅ △ ACD

We get ∠ BAD = ∠ CAD (c. p. c. t)

It can be written as

∠ BAE = ∠ CAE

Therefore, it is proved that AE bisects ∠ A.

Considering △ BDE and △ CDE

We know that BD and CD are equal sides of isosceles △ ABC

Since △ ABE ≅ △ ACE

BE = CE (c. p. c. t)

We know that DE is common i.e. DE = DE

Therefore, by SSS congruence criterion we get

△ BDE ≅ △ CDE

We know that ∠ BDE = ∠ CDE (c. p. c. t)

So DE bisects ∠ D which means that AE bisects ∠ D

Hence it is proved that AE bisects ∠ A as well as ∠ D.

(iv) We know that △ BDE ≅ △ CDE

So we get

BE = CE and ∠ BED = ∠ CED (c. p. c. t)

From the figure we know that ∠ BED and ∠ CED form a linear pair of angles

So we get

∠ BED = ∠ CED = 90^o

We know that DE is the perpendicular bisector of BC

Therefore, it is proved that AE is the perpendicular bisector of BC.