Prove that is a equilateral triangle I'm trying to find an equilateral triangle with its height given. I've done the following: Triangle
Let $C,D$ be points in the intersection of two circunferences of same radius and center $A$ and $B$. Then, I constructed the bisector of $CA$ and $DA$ and they cut in $E$ and $F$ on the circunference of center $A$ 
I have to prove that the triangle $BEF$ is equilateral but I don't know how to prove it.
 A: Since the figure drawn is symmetrical about the line AB, we have $BE = BF$ and $\alpha = 90^0$. In addition, many lines will be concurrent at one point. Examples are X, Y, and Z.

It should be clear that we only need to show $\omega + \delta = 60^0$

Since ACBD is a rhombus, $\beta = 90^0$ implies $\gamma = 90^0$ 
All red-marked angles are equal to $\theta$.
All blue-marked angles are equal to $\phi$.
Result follows from noting that (1) $\omega = 90^0 – red = 90^0 – \theta = \phi = \omega’$; and (2) $\triangle EAC$ is equilateral. .
A: I do not understand what are the givens and what you are doing in the linked figure. 
Assume that we are given $h=|AM|>0$, where $A$ is a vertex of the triangle to be constructed. Then the other two vertices $B$ and $C$ are on the perpendicular $a$ to $A\vee M$ through $M$. Draw a circle of radius $h$ with center $A$ and a second circle of the same radius with center $M$. The two circles intersect in two points $P$ and $P'$. It is well known that $\angle(MAP)=60^\circ$. Hence drop the perpendicular from $A$ to $M\vee P$ and intersect it with the base $a$. This will give you $B$; finally $C$ is obtained by reflecting $B$ in $M$.
