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Explanation:
<u>Given</u>:
- The attached figure showing circle O, chord BC, central angle BOC and inscribed angle BAC
- angle BAC = α + β
<u>Prove</u>:
- angle BOC = 2×angle BAC
<u>Proof</u>:
∠BOA +∠BOC +∠AOC = 360° . . . . . sum of arcs of a circle is 360°
2α +∠BOA = 180°, 2β +∠AOC = 180° . . . . . sum of triangle angles is 180°
∠BOA = 180° -2α, ∠AOC = 180° -2β . . . . solve statement 2 for central angles
(180° -2α) +∠BOC +(180° -2β) = 360° . . . . . substitute into statement 1
∠BOC = 2(α +β) . . . . . add 2α+2β-360° to both sides
∠BOC = 2×∠BAC . . . . . substitute given for α+β; the desired conclusion
