Circles in an Equilateral Triangle 
In the diagram below is an equilateral triangle with side length of 1 unit. $C_1$, $C_2$ and $C_3$ are circles inside the triangle tangent to each other and the sides of the triangle. Find the radius of each circle.


Firstly, I'm inclined to think that the circles must all be equal in size, but I'm not sure how to prove that. And I also tried making a smaller triangle inside the outer triangle by connecting the radii of the circles, but I'm not sure how to proceed after that (or if I'm even on the right track for that matter). 
 A: By construction you have $$1=\sqrt 3r+r+r+\sqrt 3r\Rightarrow r=\frac{\sqrt3-1}{4}\approx 0.183012701$$ Later add an explanatory figure.

A: Make a right triangle by drawing the segment connecting a vertex of your triangle to the center of the nearest circle, and dropping the perpendicular from that circle.  As this is a $30-60-90$ triangle we see that the leg along the triangle side has length $\sqrt 3 \,r$.  Inspection quickly shows that $$1=2\sqrt 3\, r+2r\implies r = \frac {\sqrt3 -1}4$$
A: Here is a simple proof that three circles, each of them touching two sides of the triangle $\triangle$  from the inside, necessarily have the same radius: 
Each circle has its center on a symmetry axis of the triangle. Given such a circle $\gamma_c$ with its center on $m_c$, and at least so large  that $\gamma_c$ intersects the incircle of $\triangle$,  the radius $r_a$ of a circle having its center on $m_a$ that touches $\gamma_c$ from the outside is uniquely determined, and due to symmetry with respect to the line $m_c$ it is equal to the radius $r_b$ of the circle having its center on $m_b$ that touches $\gamma_c$. From $r_a=r_b$ it follows that all three radii are equal.
