The relationship between facets of an inscribed polytope and those facets' shadows I posted this question thinking that the response would be two or three answers that say "Counterexamples to this are found in every textbook—for example this one and this one and this one." But the result is that after two days there are 13 views and nobody has said anything except in a private email with a comment that probably misses the mark.
Two chords of a circle of unit radius have equal lengths if their corresponding arcs have equal lengths.
Suppose a polytope is the convex hull of finitely many points on the unit $(n-1)$-sphere in $\mathbb R^n.$ Each of its $(n-1)$-dimensional facets casts a shadow on the sphere, the light-source being at the center. Suppose a proposition says two such facets have equal $(n-1)$-dimensional volumes if their shadows have equal volumes. I suspect that that is false in general, even if both facets have equally many vertices. (Unlike with chords of circles, their shapes can differ—for example not all triangles are equilateral.)
Is a proof or a counterexample known?
 A: The following is a detailed description of the counterexample given in the comments. The construction is the same, but with slightly more notation and detail for clarity. Also, the construction there gave two triangles with different areas but equal-area shadows; this answer constructs one polytope of which both of these triangles are faces (this is inessential, but it's unclear to me whether the original question asked about two facets of one polytope or of two different ones).

*

*Let $T$ be any triangle inscribed in a unit circle in the $xy$-plane. By $``$the lift of $T$ to height $\varepsilon$ in $S^2$$"$ I mean the triangle
$$\varepsilon \hat{z} + \sqrt{1-\varepsilon^2}T,$$
which is similar to $T$, inscribed in $S^2$, and lies in the horizontal plane $\{z=\varepsilon\}$. Denote this lift by $L_\varepsilon(T)$.


*If $T$ is inscribed in the unit circle in the $xy$ plane and contains the origin in the interior of its convex hull, the shadow of $L_\varepsilon(T)$ on $S^2$ approaches the entire upper hemisphere as $\varepsilon$ approaches $0$ from above, and it approaches the north hemisphere as $\varepsilon \to 1$ from below (these limits of shadows can be taken with respect to the Hausdorff metric. We care only about the areas of the shadows and not about the shadows themselves).


*Let $T_1$ be an equilateral triangle inscribed in the unit circle in the $xy$-plane and let $T_2$ be a very narrow triangle inscribed in the same circle and containing the origin in its interior, such that $T_2$ has small area $s$ (we will find a bound on $s$ later, but any $s<0.01$ will certainly do). Fix $\varepsilon_1 = 0.1$. By part (2), we can choose $\varepsilon_2>0$ such that the shadows of $L_{\varepsilon_1}(T_1)$ and of $L_{\varepsilon_2}(T_2)$ have equal area on $S^2$. It is clear that the areas of these lifts are different (the area of $L_{\varepsilon_2}(T_2)$ is much smaller, as it is bounded above by $s$, which we may modify to be smaller than $A(L_{\varepsilon_1}(T_1))$ if necessary).


*The triangles $L_{\varepsilon_1}(T_1)$ and $L_{-\varepsilon_2}(T_2)$ are inscribed in $S^2$, have shadows of equal area on $S^2$, and have different areas. The planes they lie in do not intersect. Therefore the convex hull of their six vertices is a convex polytope inscribed in $S^2$ which has them as faces. This is a counterexample.
