A way to directly see the Inscribed angle theorem? (i.e. central angle is twice the inscribed angle)

Question

Dotted line is parallel to the diameter; radii are equal length.

The upper angle is the inscribed angle; the double angle at the center is the central angle.

1. The upper triangle is isosceles, because radii are equal; therefore its base angles are equal.

2. The leftmost angle (with the dotted line) equals the inscribed angle, because they are alternate interior angles.

3. The double angle on the left equals the central angle, again by alternate interior angles.

Is there a better way to see this?

Note: This special case works whether the central angle is acute (as shown) or obtuse; it also works if reflected to the other side.

1. The first generalization is to when the arms on either side of the diameter shown is to use this construction on both sides, and add all the angles.

2. The second generalization is to where the arms are both on the same side of the diameter shown, and to subtract the lesser angles from the greater.

There's more details of the generalizations at Khan Academy's High School Geometry's Inscribed angle theorem proof ( review text; video ), but they use algebra for the special case/lemma, whereas the above is geometrical.

I really liked their reuse of the special case as a lemma - very elegant. But hard to visualize the complete proof of the special case, because it relied on triangle internal angles summing to $180^\circ$. With the great answers to this question, especially @EthanBolker's, I got a way to visualize it - which surprisingly also lead to a simpler proof, by not needing the angle supplementary to the central angle.

(I couldn't find a way to label the parallels/radii with arrows/strike-throughs at geogebra, though an otherwise excellent tool).

Answer 1

Euclid used your figure, but without the parallel, and in the more general form you mention in Note 1, to prove the theorem in question in Elements, III, 20. So you're in good company. He relied on the equal angles in an isosceles triangle, and the fact that the exterior angle of a triangle is equal to the sum of the opposite interior angles, shown in I, 32. The latter used I, 29, which rests on (parallel) Postulate 5, so @Blue's comment is apt. A more direct showing may be hard to find.

Answer 2

This illustration of Edward Porcella's answer integrates I, 32 into the first part of III, 20.