Minimizing sum of distance in a tetrahedron 
Given that $P$ is a point inside a tetrahedron with vertices at $A, B, C$ and $D$, such that the sum of the distances $PA+PB+PC+PD$ is a minimum, show that the two angles $\angle APB$ and $\angle CPD$ are equal and are bisected by the same straight line. What other pair of angles must be equal?
My progress: i considered the triangle case scenario and from there i observed angles each make with each other is 120° , now i considered a quadrilateral here i considered making two triangles and observing that the point is lying between the two points which minimize the sum in there . But this tetrahedron i am not getting how to solve like i am considering triangles for lower cases ? Or is there a better method ?

 A: Consider changing sum $S=AP+BP+CP+DP$ with changing vector $\vec{AP}$.
Let mark as vector $\frac{dS}{d\vec{AP}}$ vector with components $(\frac{dS}{dAP_x};\frac{dS}{dAP_y};\frac{dS}{dAP_z})$. For point P making $S$ minimal $\frac{dS}{d\vec{AP}}$ must be zero. $$\frac{d(AP)}{\vec{AP}}=\frac{\sqrt{AP^2}}{\vec{AP}}=\frac{1}{2\sqrt{AP^2}}\frac{d(AP^2)}{d\vec{AP}}=\frac{1}{2AP}\cdot 2\vec{AP}=\frac{\vec{AP}}{AP}$$
$$\frac{d(BP)}{\vec{AP}}=\frac{\sqrt{BP^2}}{\vec{AP}}=\frac{1}{2\sqrt{BP^2}}\frac{d(BP^2)}{d\vec{AP}}=\frac{1}{2BP}\cdot \frac{d((\vec{AP}-\vec{AB})^2)}{d\vec{AP}}=\frac{1}{2BP}\cdot 2(\vec{AP}-\vec{AB})=\frac{\vec{BP}}{BP}$$
$$\frac{dS}{d\vec{AP}}=\frac{\vec{AP}}{AP}+\frac{\vec{BP}}{BP}+\frac{\vec{CP}}{CP}+\frac{\vec{DP}}{DP}=0$$
$$\frac{\vec{AP}}{AP}+\frac{\vec{BP}}{BP}=-\left(\frac{\vec{CP}}{CP}+\frac{\vec{DP}}{DP}\right)$$
This claim can be used to solve both parts of problem.

*

*Square both sides:

$$2+2\frac{\vec{AP}\cdot \vec{BP}}{AP\cdot BP}=2+2\frac{\vec{CP}\cdot \vec{DP}}{CP\cdot DP}$$
$$2+2\cos APB=2+2\cos CPD\Rightarrow \angle APB=\angle CPD$$


*Vector $-\left(\frac{\vec{AP}}{AP}+\frac{\vec{BP}}{BP}\right)$ has the same direction as bisector of angle $APB$, and vector $-\left(\frac{\vec{CP}}{CP}+\frac{\vec{DP}}{DP}\right)$ has the same direction as bisector of angle $CPD$, so these two bisectors has opposite directions, therefore they line on the same straight line.

A: In case the tetraedron $ABCD$ is convex it is enough to prove that the requested point $P_0$ is exactly the point where the diagonals $AC, BD$ meet. It is easy to prove that using elementary geometry, as follows: Take any point $P'$ that does not lie on the segment $AC$ then by triangle inequallity we have: $AP'+CP'+DP'+BP'>AC+DB=AP_0+BP_0+CP_0+DP_0$. So  $P_0$ should lie on the segment $AC$. Similar arguments show that any point $P'$ not lying in segment $BD$ gives larger sum of distances from the vertices. So $P_0$ is the desired point. The equality of angles follows immediatelly.
