Two angles in circle produced by the arc BC are ∠BAC and ∠BDC.

∴ ∠BAC = ∠BDC …… (1)

Again, two angles in circle produced by the arc CE are ∠CAE and ∠CDE.

∴ ∠CAE = ∠CDE …… (2)

Now, ∠FDE + ∠CDE + ∠BDC = 1 straight angle = 180°.

or, ∠CDE + ∠CDE + ∠BDC = 180° [∵ DE is bisector of ∠FDC : ∠FDE = ∠CDE]

or, 2∠CDE + ∠BDC = 180° …… (3)

Again, (∠GAE + ∠CAE + ∠BAC) = 1 straight angle = 180° …… (4)

From (3) and (4) we get, 2∠CDE + ∠BDC = ∠GAE + ∠CAE + ∠BAC

or, 2∠CDE = ∠GAE + ∠CAE (from (1))

or, 2∠CAE = ∠GAE + ∠CAE (by the figure)

or, ∠CAE = ½ ∠GAC

∴ AE is the bisector of ∠GAC. Hence AE is the external bisector of ∠BAC.