Coplanarity of 4 tangent points to a sphere. Suppose that segments AB, BC, CD and DA are tangents to a sphere. I need to prove that the four tangent points are coplanar.
I really appreciate any approach or solution.
 A: Let $P$, $Q$, $R$ and $S$ be touching points of segments $AB$, $BC$, $CD$ and $AD$ to the sphere respectively.
Thus, $$AP=AS,$$ $$BP=BQ,$$ $$CR=CQ$$ and $$DR=DS.$$
Now, let the plane $(PQR)\cap AD=\{S'\}$ and $AA'$, $BB',$ $CC'$ and $DD'$ be perpendiculars to $(PQR).$
Thus, by similarity we obtain:
$$\frac{DS'}{S'A}=\frac{DD_1}{AA'},$$
$$\frac{AP}{PB}=\frac{AA'}{BB'},$$
$$\frac{BQ}{QC}=\frac{BB'}{CC'}$$ and
$$\frac{CR}{RD}=\frac{CC'}{DD'},$$ which says
$$\frac{DS'}{S'A}\cdot\frac{AP}{PB}\cdot\frac{BQ}{QC}\cdot\frac{CR}{RD}=1.$$
Can you end it now?
A: COMMENT:  Consider an arbitrary axis passing the center of sphere S. Also n planes $P_i$ parallel with this axis and tangent to the sphere. These planes intersect at lines $L_i$ with their  adjacent planes. The radii $R_i$ of sphere connecting the center and the touching points of tangent planes are perpendicular to planes $P_i$, also to arbitrary axis as the result. Hence these radii are in a plane it's normal is the arbitrary axis, let's denote it as $P_a$ . This plane intersect the planes $P_i$   and constructs lines which have following specifications:
-Their ends $V_t$ is the cross section of plane $P_a$ and lines $L_i$.
-they are tangent to the sphere S.
-They make a closed n-gonal polygon which is located in $P_a$ and has vertices $V_t$.
-They are located in one plane $P_a$ perpendicular to arbitrary axis i.e the axis is it's normal, because they all have one common normal which is the arbitrary axis.
Now the four lines you mentioned make a closed qua-dragon it's sides tangent to  sphere; this provides the fourth  specification, i.e their location in one plane,. 
