A problem about equilateral triangle In below equilateral triangle $ABC$, $AD=CE$, $DH=GH$. Prove that $BE=BG$
My thoughts: It looksl ike we need to prove that $ABGD$ are on the same circle but I am not sure how to use the condition $DH=GH$

 A: I'll prove it the other way around.
Suppose we have $AD = CD$. We find the point $G$ on the line $AE$ such that $BE = BG$, and connect $D$ and $G$. Let $H$ be the intersection of $DG$ and $CF$. I'm going to show that $DH = GH$.
Since $\angle BGE = \angle BEG = \angle AEC = \angle ADB$, we have $\angle GBF = \angle FAD = \angle ABF$, which means that $BF$ is the bisector of $\angle ABG$. Therefore we have $\frac{AB}{BG} = \frac{AF}{FG}$.
But we have also $AB = AC$ and $BG = BE = CD$, and Menelaus theorem w.r.t. the triangle $AGD$ and the line $FHC$ says $\frac{AC}{CD} \cdot \frac{DH}{HG} \cdot \frac{GF}{FA} = 1$.
This then gives $DH = GH$.
A: Let $A,B,C$ be vectors from $O(0,0)$ to $A,B,C$ resp, let $a=A-C,b=B-C$. And let $C$ be $(0,0)$ then.
Let's write all we have:
$D=(1-t)a, E=tb$,
$F$:$F=AE\cap BD$, $AE:ua+(1-u)tb$, $BD:vb+(1-v)(1-t)a$
$$\begin{cases}
u=(1-t)(1-v)\\\ 
(1-u)t=v
\end{cases}$$
$$u=(1-t)(1-t+ut)$$
$$u-t(1-t)u=(1-t)^2$$
$$u = \frac{(t - 1)^2}{t^2 - t + 1}, v = \frac{t^2}{t^2 - t + 1}$$
$$F=\frac{(t - 1)^2}{t^2 - t + 1}a+\frac{t^2}{t^2 - t + 1}b$$
$H=\frac{w}2F$
$G=xA+(1-x)E=xa+(1-x)tb$, and, finally $DH=HG$:
$2H=D+G=(1-t+x)a+(1-x)tb$
$$\begin{cases}
1-t+x=w\frac{(t - 1)^2}{t^2 - t + 1}\\\
(1-x)t=w\frac{t^2}{t^2 - t + 1}
\end{cases}$$
$$w = 2-t,x=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1},
1-x=-\frac{t (t - 2)}{t^2 - t + 1}$$
$$\hbox{Thus }G=xa+(1-x)tb=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1}a-\frac{t^2 (t - 2)}{t^2 - t + 1}b,$$
$$BE=E-B=(1-t)b, BG=G-B=\frac{2 t^2 - 3 t + 1}{t^2 - t + 1}a-\frac{(t - 1)^2 (t + 1)}{t^2 - t + 1}b$$
$$BG^2=(k_1a+k_2b)^2=k_1^2a^2+2k_1k_2ab+k_2^2b^2,$$
then recall $a^2=b^2$, $ab=|a|\cdot|b|\cos\angle ACB=\frac12a^2,$ thus $BG^2=\left(k_1^2+k_1k_2+k_2^2\right)a^2$ $=(1-t)^2a^2$ $=BE^2$, QED.
