Prove following points lie on a circle. I found this in a textbook without a solution and I wasnt able to solve it myself.
Let ABCD be a tetrahedron with all faces acute.
Let E be the mid point of the longer arc AB on a circle ABD.
Let F be the mid point of the longer arc BC on a circle BCD.
Let G be the mid point of the longer arc AC on a circle ACD.
Show that points D,E,F,G lie on a circle.
My approach to that was to try to show these point were
co-planear. Since they all lie on one sphere (the one with inscribed tetrahedron ABCD) that would solve the problem.
Needles to say I failed at that.
 A: Let $\vec{a} = \vec{DA}, \vec{b} = \vec{DB}, \vec{c} = \vec{DC}$ and $a,b,c$ be corresponding magnitudes.
Let us look at what happens on the plane holding circle $ABD$. 
Let $X$ be the circle's center and $E'$ be the mid point of the shorter arc $AB$.
It is not hard to see
$$\angle E'DB = \frac12 \angle E'XB = \frac12 \angle AXE' = \angle ADE'$$
This implies $DE'$ is the angular bisector of $\angle ADB$ and $\vec{DE'}$ is pointing along the direction $\frac{\vec{a}}{a} + \frac{\vec{b}}{b}$. Since $DE$ is perpendicular to $DE'$, $\vec{DE}$ is pointing along the direction $\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$.
To proceed, we will re-express this fact in terms of barycentric coordinates.
For any $P \in \mathbb{R}^3$, the barycentric coorindates of $P$ with respect to tetrahedron $ABCD$ is a 4-tuple $(\alpha_P, \beta_P, \gamma_P, \delta_P)$ which satisfies:
$$\alpha_P + \beta_P + \gamma_P + \delta_P = 1\quad\text{ and }\quad 
\vec{P} = \alpha_P \vec{A} + \beta_P \vec{B} + \gamma_P \vec{C} + \delta_P\vec{D}$$
In particular, the barycentric coordinates for $D$ is $(0,0,0,1)$. 
Let's look at point $E$. Since $E$ lies on the plane holding $ABD$, $\gamma_E = 0$. Since $DE$ is pointing along the direction $\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$, we find
$\alpha_E : \beta_E = \frac1a : -\frac1b$. From this, we can deduce
there is a $\lambda_E$ such that
$$(\alpha_E, \beta_E, \gamma_E, \delta_E) = \left(\frac{\lambda_E}{a}, -\frac{\lambda_E}{b}, 0, 1 + \lambda_E\frac{a - b}{ab}\right)$$
By a similar argument, we can find $\lambda_F$ and $\lambda_G$ such that
$$(\alpha_F, \beta_F, \gamma_F, \delta_F) = \left(0,\frac{\lambda_F}{b}, -\frac{\lambda_F}{c}, 1 + \lambda_F\frac{b - c}{bc}\right)\\
(\alpha_G, \beta_G, \gamma_G, \delta_G) = \left(-\frac{\lambda_G}{a},0,\frac{\lambda_G}{c}, 1 + \lambda_G\frac{c - a}{ca}\right)$$
In terms of barycentric coordinates, $D,E,F,G$ are coplanar when and only when
following determinant evaluates to zero.
$$\mathcal{D}\stackrel{def}{=}\left|\begin{matrix}
\alpha_E & \beta_E & \gamma_E & \delta_E\\
\alpha_F & \beta_F & \gamma_F & \delta_F\\
\alpha_G & \beta_G & \gamma_G & \delta_G\\
\alpha_D & \beta_D & \gamma_D & \delta_D
\end{matrix}\right|
=
\left|\begin{matrix}
\alpha_E & \beta_E & \gamma_E & \delta_E\\
\alpha_F & \beta_F & \gamma_F & \delta_F\\
\alpha_G & \beta_G & \gamma_G & \delta_G\\
0 & 0 & 0 & 1
\end{matrix}\right|
= \left|\begin{matrix}
\alpha_E & \beta_E & \gamma_E\\
\alpha_F & \beta_F & \gamma_F\\
\alpha_G & \beta_G & \gamma_G\\
\end{matrix}\right|
$$
Substitute above expression of barycentric coordinates of $E,F,G$ into last determinant,
we find
$$\mathcal{D} = \lambda_E\lambda_F\lambda_G
\left|
\begin{matrix}
\frac1a & -\frac1b & 0\\
0 & \frac1b & -\frac1c\\
-\frac1a & 0 &\frac1c
\end{matrix}
\right|
= 0
$$
as the rows of determinant on RHS sum to zero. 
From this, we can conclude $D, E, F, G$ are coplanar. Since $D, E, F, G$ lie on the intersection of a sphere and a plane, they lie on a circle.
A: I'll write the tetrahedron as $OABC$, with $O$ at the origin, and I'll let $D$, $E$, $F$ be the new points associated with faces $\triangle OBC$, $\triangle OCA$, $\triangle OAB$. Define
$$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad \alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma = \angle AOB$$
and recall that, for instance,
$$B\cdot C = b c \cos\alpha  \qquad |B\times C| = b c \sin\alpha$$

Consider the situation with $\triangle OBC$. The defining arc property for $D$ indicates that this point lies on the perpendicular bisector of $\overline{BC}$ within the plane of $\triangle OBC$. Thus, $\overrightarrow{DD^\prime}$ is perpendicular to both $\overrightarrow{BC}$ and the normal to the plane (that is, $B\times C$). we can write
$$D=D^\prime + |DD^\prime| \frac{( C-B )\times ( B \times C )}{|BC|\,|B\times C|} \tag{1}$$
where $D^\prime := \frac12(B+C)$ is the midpoint of $\overline{BC}$.
Further, by the Inscribed Angle Theorem, $\angle BOC\cong\angle BDC$, and we conclude that $\overline{DD^\prime}$ is the altitude of an isosceles triangle with vertex angle $\alpha$ and base $|BC|$. Therefore, $|DD^\prime| = \frac12|BC|\,\cot\frac12\alpha$, so we have
$$\frac{|DD^\prime|}{|BC|\,|B\times C|} = \frac{\cot\frac12\alpha}{2bc\sin\alpha} = \frac{\cos\frac12\alpha\,/\,\sin\frac12\alpha}{4bc\sin\frac12\alpha\cos\frac12\alpha}=\frac{1}{4bc\sin^2\frac12\alpha} = \frac{1}{2bc(1-\cos\alpha)} \tag{2}$$
Further, via a cross product identity,
$$\begin{align}
(C-B)\times(B\times C) &= \phantom{-}B\,((C-B)\cdot B) - C\,((C-B)\cdot C) \\
&=\phantom{-}B\left(B\cdot C - |B|^2\right)-C\left(|C|^2 - B\cdot C\right) \\
&=-B\,b(b-c\cos\alpha) - C\,c(c-b\cos\alpha)
\end{align}\tag{3}$$ 
Altogether, this gives us
$$\begin{align}D\;2bc(1-\cos\alpha) &= (B+C)bc(1-\cos\alpha)-B\,b(b-c\cos\alpha)-C\,c(c-b\cos\alpha) \\
&=Bb(c-b)+Cc(b-c) \\
&=(c-b)(B b-C c) \tag{4a}
\end{align}$$
(Note that, if $b=c$, then $D=0$, as we would expect. This confirms that we got our cross-product vector directions correct in $(1)$.) Likewise,
$$\begin{align}
E\,2ca(1-\cos\beta)  &= (a-c)(C c-A a) \tag{4b} \\
F\,2ab(1-\cos\gamma) &= (b-a)(A a-B b) \tag{4c}
\end{align}$$
Finally, observe that, for $a$, $b$, $c$ not all equal (the only case with which we need concern ourselves),
$$(\text{eq }4a)\;(a-c)(b-a) + (\text{eq }4b)\;(c-b)(b-a)+(\text{eq }4c)\;(a-c)(c-b) \tag{5}$$
gives a non-trivial linear combination of $D$, $E$, $F$ that vanishes. Consequently, $D$, $E$, $F$ are linearly dependent; that is, they lie on a common plane through $O$, and the result follows. $\square$
