ABCD is a cyclic quadrilateral. Chord DE is the external bisector of ∠BDC. Prove that AE (or produced AE) is the external bisector of ∠BAC.
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.
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