<Tri-2025-2228>, hint

Let E on the line AB such that BE=BC

Since <B=2<C,   <BEC=<BCE=<C .Triangles AEC and ACB are similar

AC/AE=AC/AB+BC=AB/AC.

In triangle AEB , AC/sinC=(AB+BC)/sin2C=(AB+BC)/2sinCcosC

2AC cosC=(AB+BC) ---> 2AC>AB+BC

<Tri-2025-2228>, general, bisector

<Romania District Olympiad 2006>

 

<Solid-2025-260>, hint 

Romania District Olympiad 2005>

Consider a box ABCDA'B'C'D' and a tetrahedron A'BC'D made of  6 diagonals of the box.

You can find 4 congruent triangular faces.

<Solid-2025-260>, tetrahedron,congrent

<Romania District Olympiad 2005>

 

<Tri-2025-2227>, hint