A curious property of an acute triangle Many years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:

$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.
The property is that $AF = AG$
Proof:
Let $H$ be the midpoint of $AB$ (and the centre of the respective semicircle)
$$AG^2 = AD^2 + GD^2 = \left(AC\cdot \cos\angle A\right)^2 + GD^2$$
Since $HG = AH = \frac{AB}{2}$ is the radius of the semicircle
$$GD^2 = HG^2 - HD^2 = \left(\frac{AB}{2}\right)^2 - \left(\frac{AB}{2} - AC\cdot\cos\angle A\right)^2 = \\ = AB\cdot AC\cdot \cos\angle A - \left(AC\cdot\cos\angle A\right)^2$$
which gives
$$AG^2 = AB\cdot AC\cdot \cos\angle A$$

Analogously ($I$ is the midpoint of $AC$)
$$AF^2 = AE^2 + FE^2 = \left(AB\cdot \cos\angle A\right)^2 + FE^2$$
$$FE^2 = FI^2 - EI^2 = \left(\frac{AC}{2}\right)^2 - \left(AB\cdot \cos\angle A - \frac{AC}{2}\right)^2 = \\ = AC\cdot AB\cdot \cos\angle A - \left(AB\cdot \cos\angle A\right)^2$$
which finally gives
$$AF^2 = AC\cdot AB\cdot \cos\angle A$$

Questions:


*

*Is this a known property?

*Is there a better more elegant proof?


HINT:
$$AF^2 = AE \cdot AC\\
AG^2 = AD \cdot AB \\
AE \cdot AC = AD \cdot AB $$
