A conjecture about the sum of the areas of $3$ triangles built on the sides of any triangle (by means of centroid/orthocenter)

Question

Given any triangle $\triangle ABC$, let us draw its orthocenter $D$. By means of this point, we can draw three circles with centers in $A,B,C$ and passing through $D$.

These circles intersect in the points $E,F,G$, which can be seen as the vertices of three triangles $\triangle AFC, \triangle CGB$ and $\triangle BEA$.

My conjecture is that

The sum of the areas of the triangles $\triangle AFC, \triangle CGB, \triangle BEA$ is equal to the area of the triangle $\triangle ABC$.

Furthermore,

If we substitute the orthocenter $D$ with the centroid of $\triangle ABC$, then the areas of $\triangle AFC, \triangle CGB$ and $\triangle BEA$ are all equal, and their sum is equal to the area of the triangle $\triangle ABC$.

Maybe these are very well known theorems. However, is there a compact proof for such conjectures?

NOTE: These conjectures are very similar to the one exposed in this post.

Thanks for your help, and sorry for imprecision or triviality.

Answer 1

This is true since for any point in triangle (not just centroid or orthocenter):

$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)

$ABE$ is congruent to $ABD$ (sss) and

$BCG$ is congruent to $BCD$ (sss).

Answer 2

Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle. So the conjecture is correct.