**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.

**Proof: **

∠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)