construct triangle with $\hat C$ and length of the bisector of $\hat C$ and side c I we know $\hat C$ and length of the bisector of $\hat C$ and side c then  how can we construct our triangle?
my attempt:
if $\hat C = \alpha$, now I draw this.

each point that I choose on two arcs (point C) and connect it to A and B, make triangle ABC with angle $\hat C = \alpha$ and AB = c. But I need some of points that make bisector angle $\hat C$ with my own length of the bisector of $\hat C$. How can I find that?  
Update 1: 
Hi. Thanks RicardoCruz for answer. I get some help from my teacher. He said:  

we assume that we draw our $\triangle ABH$. now we construct bisector of $\angle H$ that intersect with arc in G. G is in middle of arc AGB. so for construct our real triangle we can find middle of this arc then we make circle with center of G and radius $(n + x)$ to find place of point H.  so only thing that we must find length of $x$. He said that you can use similarity of triangle.
I use this way in picture so we have a quadratic equation and we can find $x$. can you have any other better way for find $x$?
I'm sorry for bad English. Thanks.
 A: Let $\triangle ABC$ such that $m(\angle ACB)=\gamma$, $S$ is the intersection point between the angle bisector of $\angle ACB$ and side $AB$, $CS=w$, $AB=c$, and $m(\angle CSB)=\theta$.
Using the sine rule in $\triangle ACS$ and $\triangle BCS$, we get:
$$SB=w  \frac{\sin{\frac{\gamma}{2}}}{\sin(\theta+ \frac{\gamma}{2})} \quad(1) $$
and
$$AS=w  \frac{\sin{\frac{\gamma}{2}}}{\sin(\theta- \frac{\gamma}{2})}. \quad(2) $$
Recalling that $AS+SB=c$ and after some algebra with trigonometric identities, we get:
$$-\frac{c}{w \sin \gamma}\sin^2(\theta)+\sin(\theta)+\frac{c}{2w}(\frac{1}{\sin \gamma}-\frac{1}{\tan \gamma})=0. \quad (3)$$
Solving for $\sin\theta$ the quadratic equation $(3)$, we can easily construct the $\triangle ABC$.
A: EDIT: It turns out that this post has an error, namely that one of my algebraic manipulations is incorrect.  (I knew this was too good to be true....)  I haven't been able to modify the solution to make it still work because now the degrees of the relative terms don't yield to the same techniques, but I still leave the solution up hopefully to serve as inspiration for other possible solutions.
For ease of typesetting, let $\triangle ABC$ be the triangle in question, and let $d$ be the length of the angle bisector from $A$.  Note that by the angle bisector theorem it's easy to prove that this bisector divides $BC$ into two segments of length $\frac{ac}{a+b}$ and $\frac{bc}{a+b}$.
Now recall Stewart's Theorem, which states that whenever $D$ is a point on $BC$, if $BC=a$, $AC=b$, $AB=c$, $AD=d$, $BD=m$, and $CD=n$, the relationship $$b^2m+c^2n=d^2a+amn$$ holds.  (The easiest way to remember this is to use the mnemonic "A man and his dad put a bomb in the sink" :) ) Plugging this into the result in question gives $$a^2\left(\dfrac{bc}{a+b}\right)+b^2\left(\dfrac{ac}{a+b}\right)=d^2c+c\left(\dfrac{ac}{a+b}\right)\left(\dfrac{bc}{a+b}\right).$$ Note that the left hand side simplifies to $abc$, so dividing through by $c$ yields $$ab=d^2+\dfrac{abc^2}{(a+b)^2}.$$ Now some algebraic manipulation gives $$d^2=\dfrac{ab(1-c^2)}{(a+b)^2}\implies \dfrac{1-c^2}{d^2}=\dfrac{(a+b)^2}{ab}=\dfrac ab+\dfrac ba + 2.$$  This is actually a quadratic in $a/b$, so we can solve for the ratio $a/b$.  This is actually great news, because the segment $AD$ divides $BC$ into two segments whose lengths are in the ratio $a/b$!
With this in mind, I propose the following construction:


*

*Construct the angle $C$.  Extend the sides of $\angle C$ far out with a straightedge.

*Construct the angle bisector of $\angle C$.  Extend it to a point $D$ so that the length of $CD$ is $d$ (which is already given to you).

*Pick an arbitrary point $P$ on one of the sides of $\angle C$.  Now construct a point $Q$ on this same side so that $CP/CQ=a/b$.  (This is the part which I'm a bit unsure about, but considering that $a/b$ can be expressed entirely in terms of the lengths of $c$ and $d$ combined with nothing more complicated than square roots and squares, I'm pretty sure there exists a routine but tedious algorithm to do this.)

*Construct a circle with center $C$ and radius $CQ$.  This intersects the other side of the angle $C$ at some point $R$.

*Draw a line through $D$ parallel to $PR$.  Let it intersect the sides of $\angle C$ at $A$ and $B$.

*Stare at your newfound creation, which should be $\triangle ABC$.

