The perimeter of the triangle \(ABC\) is 10. Let \(D,E,F\) be the midpoint of the segments
\(AB,BC,AC\) respectively. What is the
perimeter of the triangle \(DEF\)?
Let \(ABC\) be a triangle and \(D\) be a point on the edge \(BC\) so that the segment \(AD\) bisects the angle \(\angle BAC\). Show that \(\frac{AB}{BD}=\frac{AC}{CD}\).