[Thales semicircle theorem](http://en.wikipedia.org/wiki/Thales%27_theorem) says that an angle inscribed in a semicircle is a right angle.

> ***Q1***. Does a cone with apex on a hemisphere and encompassing the circular base
have a solid angle independent of the position of the apex?

It appears it *might* be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:
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&nbsp;
![ConeSolidAngle][1]
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> ***Q2***. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be *No*...
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**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will,
let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection
of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$
left to right:
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&nbsp;
![ConeSolidAngle3][2]
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And here are two enlarged views of the $\phi=5^\circ$ intersection:
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&nbsp;
![ConeSlid5degx2][3]
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The area of the (blue) spherical polygon is the solid angle.
As Will argues, its area approaches $\frac{1}{4} (4 \pi)=\pi$ as $\phi \to 0$, so the answer to ***Q1*** is *No*.


  [1]: https://i.sstatic.net/FepAS.jpg
  [2]: https://i.sstatic.net/5j0y5.jpg
  [3]: https://i.sstatic.net/U3Wnz.jpg