There is a class 9 or class 10 theorem that "The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle." Now the proof which is taught to us is like-
Let there be a circle with centre O.
$$∠BOQ=∠OAQ+∠AQO …..(1)
$$Also, in △ OAQ,
$$OA=OQ [Radii of a circle]
$$Therefore,
$$∠OAQ=∠OQA $$ [Angles opposite to equal sides are equal]
$$∠BOQ=2∠OAQ …….(2)
$$Similarly, BOP=2∠OAP ……..(3)
Adding 2 & 3, we get,
$$∠BOP+∠BOQ=2(∠OAP+∠OAQ) $$
$$∠POQ=2∠PAQ …….(4) $$
This I have understood. But let's say I altered the diagram like this.
In this O is the centre I tried using exterior property but failed. Mind you you might start proving it by same segment. Angle in same segment are equal. If we using this segment property then we can prove this but to prove that angle in same segment are equal we use angle subtended in circle is double at any point on circle and like using each other to prove each other seems a bit wrong. So is really the angle subtended at centre is wrong or I am missing to solve it here?
Best Answer
The standard approach to the Inscribed Angle Theorem, as shown in most high school textbooks on geometry, is to start by proving it for the special case when one side of the angle is a diameter. They then treat the general case by drawing in an auxiliary diameter through the vertex of the angle, as shown below. Case 1 is from the source, and Case 2 is from the OP's drawing here.
Edit: At the original poster's request, I am providing an explicit proof for Case 2, rather than an argument sketch.
As in Case 1, the proof begins by drawing the diameter $AB$, showing that the measure of Angle $PAB$ is half the measure of Arc $PB$, and that the measure of angle $QAB$ is half the measure of Arc $QB$. As these proofs were given in the original post; as the arguments are identical in both cases; and as drawing in the radii would needlessly complicate the diagrams, I trust it is safe to omit full demonstrations of these preliminaries.
Once we've established that $m \angle PAB = \frac{1}{2} m \text{ Arc } PB$, $m \angle QAB = \frac{1}{2} m \text{ Arc } QB$, the proof of Case 2 looks a lot like Case 1, except we subtract instead of add: since $$m \angle PAQ = m \angle PAB - m \angle QAB$$ and $$m \text{ Arc } PQ = m \text{ Arc } PB - m \text{ Arc } QB,$$ we get
\begin{align*} m \angle PAQ &= m \angle PAB - m \angle QAB \\ &= \frac{1}{2} (m \text{ Arc } PB - m \text{ Arc } QB) \\ &= \frac{1}{2} m \text{ Arc } PQ, \ \end{align*} and our argument is complete.
Since you ask for a "reputable source," here is how Pearson's Common Core geometry textbook, widely used in US high schools, approaches the Inscribed Angle Theorem (pp. 780-781):
The pictures they draw for Exercises 26 and 27, in the same section of the textbook, are substantially equivalent to the ones I drew above for Cases 1 and 2 above (which the textbook refers to as "Case II" and "Case III" respectively). It seems noteworthy, per OP's request for a proof, that the student edition of the textbook does not explicitly fill in the proof of Cases 1 and 2 above, but assumes that a motivated and intelligent high school student would be able to come up with the details themselves, from seeing the demonstration when one side of the angle is a diameter.
Edit 2: Wanted to provide another source. Matthew Harvey (math professor at UVA) has a fascinating and readable free geometry textbook available on his university webpage. He gives an explicit proof of the Inscribed Angle Theorem on pages 12-13 of Chapter 16, "Circles". Harvey's proof is exactly the one given in my answer, and outlined in Pearson's geometry textbook. It covers the case which OP was confused about, where angle/arc subtraction is necessary.