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Angle in the alternate segment

In the diagram, a chord BD is drawn to the point of contact of the tangent DC. Click and drag any of the blue points to change the shape and orientation of the figure. What do you observe?

Let <BDC = α°.

1. Why is <ODC = 90°?
2. From (1) we can say that <ODB = 90° - α. Why is <OBD = 90° - α?
3. Also, <DOB = 180° - 2(90° - α) = 2α. Why is this so?
4. Given the result from (3), <BAD = α. Why is this so?
5. What does the result from (4) prove?

© State of New South Wales, Department of Education, 2018