How to prove that in any triangle there is this point X. i am having difficulties in proving this assertion:

In any triangle there is a point $X$ which is the point of
  intersection of three circles of equal radii,each of which is
  internally tangent to two sides of a given triangle.

So far i wasn't able to prove this...can someone help ? 
 A: It would be fair to split the answer into two parts: general proof of the theorem and constructional solution for a given triangle.
Note also, that it probably depends on the person who accepts the proof, so it can come to disputable situation, whether the proof is strict and full enough. I personally would accept the following proof, since it is logical and supplied with constructional example.  
So consider a circle with the point $X$ (center) and three arbitrary points $A_0$, $B_0$, and $C_0$ on its circumference. Consider now three circles of same radii, which intrsect in point X (obvious fact), and build (circumscribe?) a triangle $ABC$ around these circles.

Now, most important point: this operation on the set of possible points ($A_0$, $B_0$, and $C_0$) generates the set of all possible triangles, which includes of course also our triangle $ABC$ as its element. Paraphrased in other direction: it means that any triange $ABC$ has this point $X$ and it is obvious that $X$ belongs to the area of this triangle which is seen from the construction: by moving the initial points along the circle, you can never come to a situation, where the point $X$ goes outside the triangle $ABC$.
So that would be a simple proof of the assesment, in my opinion. 

Now to the construction problematic. Note that probably you cannot do it without having a quite precise measurement tool (ruler with scale), because you'll need to measure and calculate the linear proportion.


*

*Build the circumscribed circle around $ABC$, which is accomlished by building perpendiculars through centers of each side (orange) and draw same circles at triangle vertices:


*Buld the big triangle $A''B''C''$ around those circles and draw bisectors of each angle. This will give a center point $O$ as intersection of bisects. Now you'll see, that you have the same upscaled triangle, and all you need now is to downscale the three circles to fit inside the $ABC$. Since you cannot do that with hand tools, you'll do a pair of measurements and calculations.


*Namely you need three points on the bisectors, which will be centers of needed circles. Since that will be a uniform scale against the point $O$, you can calculate the scale factor as relation between the size of $OA$ and $OA''$ (or any other pair, since the relation is the same for all directions): 
$$k = \frac{OA}{OA''}$$ 

*Now calculate the sizes of the needed segments, which you'll lay from the point $O$ in each direction along bisectors:
$$OA' = OA*k$$
$$OB' = OB*k$$
$$OC' = OC*k$$
So you'll get the needed centers of three circles:

A: As Mikhail V has remarked the centers of the three circles are the vertices of a triangle lying homothetic to the given triangle $ABC$, with the incenter $I$  of $ABC$ as center of the homothety.
We know of three circles of equal radius that have $A$, $B$, $C$ as centers and are meeting at a common point. They are the three circles with radius the circumradius $r$ of $ABC$; they meet at the circumcenter $M$ of $ABC$. Draw these "large" circles. There is a triangle $A'$, $B'$, $C'$ homothetic to $ABC$ as well, each of whose  sides is tangent to two of the three large circles. We now have a configuration of the desired kind, albeit with respect to the triangle $A'B'C'$. A scaling with ratio ${\rho\over\rho +r}$ ($\rho$ is the inradius of $ABC$) will then produce the desired circles in the original triangle $ABC$.

