Bisector of angle lies between height and median of triangle at a vertex I want to prove that internal bisector of angle A is ( always lies) between height and median lines of triangle ABC. 
Traingle
Is it possible to help me?
Thanks.  
 A: 
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ Proof without words :




Well, not quite. :-$)$ In order to grasp things intuitively, first fix two things: the size of the angle  $($in green$)$, and that of the angle bisector $($the segment connecting the two yellow dots$)$. Keeping  the foot of the bisector fixed $($lower yellow dot$)$, we let the other two vertices of the triangle slide  freely along the two lines which determine the given angle. Thus we have the two pairs of red  and blue points. The former determine an isosceles triangle, where all three lines $($median, bi–  sector, and height$)$ coincide. In the latter case, by moving one vertex higher, $($and thereby closer  to the upper tip of the triangle, determining the height to fall on the right half of the figure$)$, the  other one is inevitably lowered, causing the position of the median's foot to descend, by stretch–  ing the length of the left portion of the triangle's base $($uniting the lower-left blue dot to the foot  of the bisector$)$, while simultaneously shrinking its right side $($uniting the foot of the bisector to  the upper-right blue dot$)$.
A: Among the three, the altitude (AH) is definitely the shortest because it is the perpendicular distance of A from BC.
In your figure, we can assume that AC > AB.
By angle bisector theorem, AC : AB = CD : DB. This means CD > DB.
Also, from the fact that M is the midpoint of BC, we can say M is on the right of D. 
If this is so, we have a triangle ABM with AD being an internal line of it. Then see my work in  Proof of an inequality in a triangle 
