Question 39
Consider a tetrahedron $ABCD$ with $\widehat{BAC} + \widehat{CAD} + \widehat{DAB} = 180^\circ$ and $\widehat{ABC} \equiv \widehat{DAB}$. If the projection of the vertex $D$ on the plane $(ABC)$ is the orthocenter of the triangle $ABC$, prove that $AB = AC$ and $DB = DC$.