Equilateral triangle ABC and two small triangles

Question

prove that $\bigtriangleup EDC \cong \, \bigtriangleup DTB$, where T is the centre of gravity and $\bigtriangleup$ ABC is equilateral. And you know that |AE| = $\frac12$|EC| and |CD| = $\frac 12$|DB|.

Thank you in advance.

Answer 1

Extend BT to meet AC at X.

Since T is the centroid, BX is the height, with BT = $\frac 23$BX. Then, the triangles BXC and BTD are similar due to BD = $\frac 23$BC and shared $\angle$CBX, which leads to

$\angle$ECD = $\angle$TDB; $\>\>\>$ TD = $\frac 23$XC = AE = CD

Together with CE = DB, the triangles CDE and DBT are congruent.

Answer 2

Let us also take the point $F$ on the third side as below:

The parallels through $D,E,F$ to the "other two sides" cut the triangle in nine smaller equilateral triangles.

(Alternatively, start with one of those small triangles, and build the picture by successive reflections in the sides.)

Three of these parallels from $D,E,F$ are passing through $T$, since $T$ divides each median segment in the proportion $1:2$, exaclty the same as the points $D,E,F$ are dividing the sides. Let us use as a unit the length of the side of each "small equilateral triangle in the picture". Then $\Delta EDC$, $\Delta DTB$ have each an angle of $60^\circ$ between the sides of lengths $1$ and $2$.