Constructing three mutually tangent congruent circles inscribed in an equilateral triangle I've been trying to construct the following figure geometrically:

I've been tearing my hair out all afternoon. Because of the irrational radius lengths of the circles, this problem is (at least to me) incredibly difficult. 
If I was given the triangle joining the centers of the congruent circles and asked to draw the circles and the large triangle, it would be easy. But given the large triangle how can I do this? Can I work it backwards somehow?
To clarify, the triangle is equilateral and the circles are tangent to each other, and to the triangle.
Please help!
 A: For an equilateral triangle $ABC$, I believe the construction using the fewest elementary operations is:


*

*Construct $CM$, the perpendicular bisector of $AB$, which will intersect $C$; $M$ is the midpoint of $AB$.

*Construct the angle bisectors of angle $CAB$ and angle $CMA$; these meat at point $P$.

*Drop a perpendicular from $P$ meeting $AB$ at $T$.

*Construct $BN$, the perpendicular bisector  of $AC$; $N$ is the midpoint of $AC$.

*Construct $Q$ on $CM$ such that $CQ = AP$, and $R$ on $BN$ such that $BR = AP$.
The centers of the three circles will be at $P,Q, R$ and the radii will be the length of $PT$.
A: $\def\c#1{\mathcal{C}_#1}\def\l#1{\mathcal{L}_#1}$
Let $a=|AB|=|BC|=|CA|$. 


*

*Find a middle point $M$ on $AB$, draw line $\l1=CM$.

*Draw a circle $\c1(M,|MA|)$, $D=\c1\cup\l1$ ($D$ is the center of the upper circle).
Let $R=|AD|$ 

*Draw three circles $\c{A}(A,R)$, $\c{B}(B,R)$, $\c{C}(C,R)$.
$\quad E=\c{B}\cup\c{C}$, $\ F=\c{A}\cup\c{C}$.

*Draw line $\l2=EF$, $\quad G=\l1\cup\l2$.
The centers of the circles in question 
are $D,E,F$; the radius $r=|EG|$.

A: To construct three circles with equal radius within an equilateral triangle, we may follow the following steps:
i) At first, we can draw a straight line of any length.
ii) Next, we have to mark a point at half of the line and the two one-fourths on both side of the half mark.
iii) Then we have to draw three perpendiculars from the three points marked on the straight line.
iv) Then, on the perpendiculars drawn from the points of the one-fourths on two sides, we have to mark a point with length one fourth of the base line i.e. the initial line.
v) Then with the radius of one-fourth part of the base line, we have to draw two circles from the two points marked on the two lines on the base line.
vi) Then on the middle line, we have to choose a suitable point from where if a circle is drawn, it would meet the two circles o the one-fourth lines only at the circumference and we can draw the circle.
vii) Then on the right side, we have to draw a tangent such that it intersects the two circles at the circumference and the same on the left side.
viii) Now, we can extend the base line on both sides (if required) to meet the two tangents.
ix) So, our equilateral triangle with three circles with same radius in it is ready.
For the sake of convenience, let us take an example:
i) At first, we can draw a line segment of 10cm.
ii) Then we have to mark points on 5cm from any side, 2.5cm from left side and 2.5cm from right side.
iii) Then we have to draw perpendiculars from the points.
iv) Then, we have to mark points of 2.5cm on the lines on left and right sides but not on the middle one.
v) Then we have to draw two circles of radius 2.5cm from the two points so that they meet each other at the circumference.
vi) Now, we can find a suitable point on the middle line so that if a circle is drawn with radius 2.5cm, it would meet the two circles only at the circumference.
vii) Then, we have to draw the circle.
viii) Then, we have to draw a tangent on the left side so that it would meet the two circles at two points on their circumference.
ix) We have to do the same procedure on the right side.
x) Then, we have to extend the base line (10cm line) if required so that it would meet the two tangents.
xi) Our equilateral triangle is ready.
