Page 8 - ln1

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 getBE = CE and ∠ BED = ∠ CED (c. p. c. t)
From the figure we know that ∠ BED and ∠ CED form a linear pair of angles
0
So we get∠ BED = ∠ CED = 90
We know that DE is the perpendicular bisector of BC
Therefore, it is proved that AE is the perpendicular bisector of BC.