enter image description here

You should also notice that since $CP$ is the diameter of the circle, angle $CBP$ is a right angle.

says the solution to this question. But what is the reasoning or proof behind the claim that the angle opposite the diameter ($\angle CBP$) must be a right angle?

I started to draw it out to gain some intuition, but I only got about as far showing that drawing a line from the center $O$ to the vertex of $CBP$ splits the inscribed triangle into two, where $\angle BOC + \angle BOP = \pi$. I'm not sure if that will lead to a proof, or whether this is more easily shown with e.g. the Law of Sines.

See also:

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    $\begingroup$ This is Thales' theorem. $\endgroup$ – Rasmus Aug 24 '13 at 7:47
  • $\begingroup$ There are various proofs here en.wikipedia.org/wiki/Thales'_theorem $\endgroup$ – Mark Bennet Aug 24 '13 at 7:54
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    $\begingroup$ The first proof on the Wikipedia page (linked in Mark Bennet's comment) does indeed start by drawing a line from the center of the circle to the opposite angle. $\endgroup$ – Rahul Aug 24 '13 at 7:57
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    $\begingroup$ I think I see now what happened: apparently you have a typo there, @Trevor, since both in that link's answers and in your own post, there appears $\,\angle CBP\;$ , yet the first time you wrote $\,\angle CPB\;$ ...In fact what you have there is that $\;PB\;$ is the height to the hypothenuse $\;AC\;$ in the straight- angled triangle $\,\Delta ACP\;$ and we know such a height divides the original triangle in two smaller one which are similar to each other and also to the original triangle. $\endgroup$ – DonAntonio Aug 24 '13 at 9:49
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    $\begingroup$ Hint: (for original question) from the knowledge that $\angle PBC$ is a right angle $\triangle ACP$, $\triangle APB$ and $\triangle PBC$ are similar. $\endgroup$ – Warren Hill Aug 24 '13 at 13:33

enter image description here

In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.

In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$


$$2 \cdot \alpha + \theta = 180 ^\circ$$

In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$


$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$

Add both these together

$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$

Which we can easily rearrange to show:

$$ \alpha + \beta = 90 ^\circ $$

We have thus proved $\angle CDB$ is a right angle as required.

  • $\begingroup$ Of course, you don't need to mention $\theta$. The picture shows straight away that the sum of the internal angles of $BCD$ is $2\alpha+2\beta$, whence $\alpha+\beta$ is 90 degrees, as required. $\endgroup$ – HJRW Apr 10 '17 at 8:13

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