Is this construction possible? Is it possible to construct in an equilateral triangle $3$ congruent circles such that they're internally tangent to two sides of the triangle and shares a point of intersection between each other?
I've been trying to make such construction but i've miserably failed so far,so i was wondering if such construction is possible in general.
If it is,how do you construct it ?
Thanks in advance.
Edit: 
It should be something like that but with the exception that the three circles must have a point of intersection in common.
 A: Let $\triangle{ABC}$ be an equilateral triangle.


*

*Get a point $D$ on the side $BC$ where $BD=\frac 13BC$. (Do you know how?)

*Draw the perpendicular line to the side $BC$ that passes through $D$. (call this line $L_1$)

*Draw an angle-bisector of $\angle{CBA}$. (call this line $L_2$)

*The intersection point between $L_1$ and $L_2$ is the center of one circle you want.
A: Here is how you can do it without perpendiculars and without getting the 1/3 of the side.


*

*Draw a circle and a same circle which goes through the center of the first one:  





*Connect the marked points with straight lines and you'll get something like this. Now you draw again the same circle (orange) at the point O and now you have automatically the centers of two other circles, which are intersections with other two circles (yellow) :  


 


*Now just draw two circles again at those centers and youre actually almost done:  




To solve the task for a given triangle, I would do it so:


*

*Construct the heights in each direction with big circles, which will also give the center of the triangle, then draw the circle around the triangle:  


 


*Connect the points from previous image and you'll get this, then draw again three lines (green) and you're done, here are the centers of the needed circles:  



