# Equivalent condition - regular tetrahedron

How can one show that if in a tetrahedron all plane angles in every vertex separately are equal, then this tetrahedron is regular?

What can I use here?

If $a$, $b$, $c$, $d$ are the plane angles about respective vertices $A$, $B$, $C$, $D$, then just we just solve the linear system \begin{align} a + b + c &= 180^\circ \qquad \text{(from } \triangle ABC \text{ )} \qquad (1)\\ a + c + d &= 180^\circ \qquad \text{(from } \triangle ACD \text{ )} \qquad (2)\\ a + d + b &= 180^\circ \qquad \text{(from } \triangle ADB \text{ )} \qquad (3)\\ b + c + d &= 180^\circ \qquad \text{(from } \triangle BCD \text{ )} \qquad (4) \end{align}
Note that subtracting any equation from any other reveals a pair of angles to be equal; in particular, \begin{align} (1)-(2) &\qquad\to\qquad b = d \\ (1)-(3) &\qquad\to\qquad c = d \\ (1)-(4) &\qquad\to\qquad a = d \\ \end{align} Therefore, $a=b=c=d = 60^\circ$, so that each face is equilateral (and necessarily congruent to the others), making the tetrahedron regular. $\square$
Let the tetrahedron have vertices $A$, $B$, $C$, and $D$. Let the angles at $A$, $B$, $C$, and $D$ be $a$, $b$, $c$, and $d$ respectively. Then because $ABC$ and $ABD$ form triangles, we have $c=d=180-a-b$. Now consider triangle $ACD$, from this we get $180=a+c+d=a+2c$, but considering triangle $ABC$ again we have $180=a+b+c=a+2c$. Hence $b=c=d$. Considering triangle $BCD$ now gives us $180=b+c+d=3c$, so $b=c=d=60$. Considering triangle $ABC$ one last time, we have $180=a+b+c=a+120$. Hence $a=b=c=d=60$, so every face of the tetrahedron is an equilateral triangle, so the tetrahedron must be regular and we are done.