Congruence of 3 Equilateral Triangles

Question

I've been having some trouble with this question from my geometry booklet, any help would be appreciated

Let ABC be an acute-angled triangle. To the outside of triangle ABC attach equilateral triangles ABD, BCE and CAF, prove that line segments AE, BF and CD all have equal length

Thank you in advance!

Edit: I have tried to learn the Menelau's Theorem in order to attempt this question but did not see any relevance, I have learnt that they all intersect at the same point but am unable to describe that in mathematical terms

Answer 1

Another approach:

$\triangle CBF=\triangle CAE$ for SAS(two sides and angle between are equal).So $AE=BF$. For the similar reason $\triangle BCD=\triangle BAE$ so $AE=CD$ , therefore $AE=BF=CD$.

Answer 2

We will prove that $BF$ and $CD$ have the same length. By symmetry, this will imply that $AE$ has equal length to the other two.

Note that clearly $B$ is a rotation of $D$ centered about $A$ by $60^\circ$ clockwise (assuming WLOG the vertices of $\Delta ABC$ are marked counterclockwise) and $F$ is a rotation of $C$ centered about $A$ by $60^\circ$ clockwise.

Hence, $BF$ is a rotation of $CD$ centered about $A$ by $60^\circ$ clockwise. Since rotations preserve length, we have that $BF=CD$.