Given: Chord AB and chord CD intersect at E in the exterior of the circle.
To prove: ∠AEC = 1/2 [m(arc AC) – m(arc BD)]
Construction: Draw seg AD.
∠ADC is the exterior angle of ∆ADE.
∴ ∠ADC = ∠DAE + ∠AED [Remote interior angle theorem]
∴ ∠ADC = ∠DAE + ∠AEC [C – D – E]
∴ ∠AEC = ∠ADC – ∠DAE ……(i)
∠ADC = 1/2m(arc AC) (ii) [Inscribed angle therom]
∠DAE = 1/2 m(arc BD) (iii) [A – B – E, Inscribed angle theorem]
∴ ∠AEC = 1/2 m(arc AC) – 1/2 m (arc BD)
[From (i), (ii) and (iii)]
∴ ∠AEC = 1/2 m(arc AC) – m (arc BD)