Constructing a triangle. I want a proof that there is one and only one possible unique triangle ABC in which base BC is of length 6 cm,angle B=60 degrees and the sum of other two sides is 9 cm.
A proof without use of trigonometry will be appreciated 
 A: Hint:
Let $A=(-3,0)$ and $B=(0,3)$. The vertex $C$ of the searched triangle is the point of intersection, with positive $y$ (this gives the unicity), of the straight line $y=\frac{\sqrt{3}}{2}(x+3)$ and the ellipse that has foci at $A$ and $B$ and has major axis $2a=9$ 
(I think that, by this image, you can also formulate a proof without the use of analytic geometry) 

A: 


*

*Draw segment $BC = 6$ and then draw circle $k$ with center $B$ and radius $9$. 

*Take the point $D$ on the circle so that $\angle \, CBD = 60^{\circ}$ (there are exactly two choices, but they lead to constructions that are symmetric mirror images of each other across the line $BC$). 

*Choose point $M$ to be the midpoint of $CD$. 

*Draw the orthogonal bisector of $CD$ and let it intersect $BD$ at point $A$.
The triangle $ABC$ is your triangle, because $AC = AD$ and thus $$BA + AC = BA + AD = BD = 9$$ The triangle $ABC$ is unique up to euclidean congruence because all the steps in the construction are unique up to congruence.   
It is immediate to generalize this construction to the case where $BC = a$, $\,\, \angle \, ABC = \beta$ and $CA + AB = d$ are arbitrary (within certain limitations).  
A: The angle at $B$ is $60°$ and the length of $BC$ is $6$ cm.
Consider placing $A$ on the ray from $B$, initially arbitrarily close to $B$, at which point $AB+AC$ is effectively also $6$ cm. As $A$ is moved away from $B$, $\angle BAC$ decreases towards a right angle and $AC$ decreases continuously, but more slowly than $AB$ is increasing, so $AB+AC$ is increasing. When $\angle BAC$ becomes a right angle, we have half an equilateral triangle and so $AB=3$ cm, with $AC<6$ cm. 
So $AC$ needs to be acute. As $A$ moves further away from $B$, both sides will now increase indefinitely. We reach the point that satisfies  $AB+AC = 9$ cm and then all more distant placements of $A$ will have $AB+AC > 9$ cm. This gives a unique positioning of $A$ on this given ray from $B$.
Arguably you have a choice of two $60°$ rays from $B$, giving two cases that are reflection of each other about $BC$.
