A curious property of an acute triangle
Solution 1:
HINT:
$$AF^2 = AE \cdot AC\\ AG^2 = AD \cdot AB \\ AE \cdot AC = AD \cdot AB $$
Solution 2:
This is a possibly similar proof, but here goes anyway...
Let the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\theta$
Then $$AF=b\cos\theta$$ $$\implies FE=AF\sin\theta=b\sin\theta\cos\theta$$ $$\implies EC=FE\tan\theta=b\sin^2\theta$$ But $$EC=b-c\cos A=b\sin^2\theta$$ $$\implies b\cos^2\theta=c\cos A$$ $$\implies AF=\sqrt{bc \cos A}$$
To get $AG$ we only need to exchange $b$ and $c$, so the result follows.