# 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.

• @MauroALLEGRANZA I want to prove why this construction is valid hahahaha – Alopiso Nov 29 '16 at 11:10

## 2 Answers

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. .

• Hi @Mick! I don't understand how you show that $\omega + \delta =60º$. Can you explain it to me? Thank you ;) – Alopiso Dec 1 '16 at 14:34
• @Alopiso $\omega + \delta = \omega' + \delta$ (proved above). But $\omega' + \delta= 60^0$ because it is an angle of $\triangle EAC$ which is equilateral (3 sides equal in length). – Mick Dec 1 '16 at 15:56
• Thank you!. One more question. Why the 2 red angles with vertex A are equal to $\theta$? Thanks a lot :)) – Alopiso Dec 1 '16 at 18:14
• @Alopiso They are equal simply by symmetry. Or they can be proved to be equal via congruent triangles. Each of them is equal to $\angle FYE$ because "angle at center is equal to 2 times the angle at circumference". – Mick Dec 2 '16 at 2:33

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 $M$.