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Prove that, if two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle.

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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)

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