Why is the green theta angle actually equal to theta for the Inscribed Angle Theorem?

Question

Source:https://en.m.wikipedia.org/wiki/Inscribed_angle

I don't understand how the green "theta" angle is claimed to be equal to "theta." The theta angle is not 90 degrees, and if green theta were equal to theta, the yellow theta would also be equal to theta, but that's a contradiction because two angles on the same line add up to 180 degrees, and their equality would imply they are equal to 90 degrees, but that's not possible per Thale's theorem only inscribed angle 90 degrees corresponds to a right triangle with hypotenuse's length equal to the diameter.

Answer 1

The yellow theta is not equal to theta: note the bit of the theorem that says "subtends the same arc": the arc subtended by the two black thetas is the same (the minor arc below the chord) but the arc subtended by the yellow theta is the major arc above the chord: the angles subtended by the major and minor arcs cut off by a chord sum to 180 degrees - as the yellow and green thetas do. The Wikipedia quote should say "Therefore, the angle does not change as its vertex is moved to different positions on the circle, provided you don't move it to the other side of the chord".

Answer 2

We can first deduce that the yellow angle is $\pi-\theta$ because it subtends an arc of measure $2\pi-2\theta$: the arc that goes the long way around the circle between the endpoints of the chord.

The green angle does not subtend any arc; one of its rays is pointing out of the circle, not even tangent. We cannot deduce its angle measure that way. But because the yellow and green angles add up to $\pi$, and the yellow angle is $\pi-\theta$, we know that the green angle is also $\theta$.

Answer 3

As shown in the diagram below,

$ABCD$ is a cyclic quadrilateral, so it's opposite angles are supplementary. Thus, $\measuredangle ADC = 180^{\circ} - \theta$ and, due to $ADE$ being a straight line, then $\measuredangle CDE = 180^{\circ} - \measuredangle ADC = \theta$.

Note with $AFCD$ being a cyclic quadrilateral, this supplementary angle property can also be used to prove that $\measuredangle AFC = \measuredangle ABC = \theta$.