Find the angle if the area of the two triangles are equal? 

Let $I$ be the incenter of $\triangle ABC$, and $D$, $E$ be the midpoints of $AB$, $AC$ respectively. If $DI$ meets $AC$ at $H$ and $EI$ meets $AB$ at $G$, then find $\measuredangle A$ if the areas of $\triangle ABC$ and $\triangle AGH$ are equal.

I played around with GeoGebra and found out that it should be $60^{\circ}$, but am unable to prove it. I saw that $\cot \measuredangle HDA= \frac{a-b}{2r}$ and some stuff like that, but its not really helping me. Can anyone solve it, preferably by pure geometry? :)

Here's my trigonometric endeavors: 
Take $\alpha=\measuredangle AID$. Using the fact that $ID$ is the median of $AIB$, we have that $\cot \frac{A}2+\cot \alpha= \cot \frac{B}2+\cot (180^{\circ}-\alpha)$, or thus: $$\cot \alpha=\frac{\cot \frac{B}2 - \cot \frac{A}2}2=\frac{(s-b)-(s-a)}{2r}=\frac{a-b}{2r}=\frac{s(a-b)}{2[ABC]}$$
So we have:
$$8[ADH]=AH\cdot 2c\sin A=\frac{c^2}{\frac{s(a-b)}{[ABC]}+2\cot A}$$
Thus, using that $2c\sin A=\frac{4[ABC]}{b}$, we have:
$$AH=\frac{c^2}{2c\sin A\cdot(\frac{s(a-b)}{[ABC]}+2\cot A)}=\frac{c^2}{2[\frac{2s(a-b)}{b} +2c\cos A]}=\frac{c^2}{\frac{2[2s(a-b)+b^2+c^2-a^2]}{b}}$$
So: $$2[2s(a-b)+b^2+c^2-a^2]=2[(c+a+b)(a-b)+b^2+c^2-a^2]=2c[a-b+c]$$
Hence, $$\frac{AH}{AC}=\frac{c}{2[a-b+c]}$$
Wherein we can continue with Jack's solution .. but I have got a extra 2? :/
