In a $ΔABC$, $D$ is midpoint of $BC$ such that $DA=DB=DC$, $BE$ and $CF$ are angle bisectors of $ABC$ and $ACB.EG$ and $FH$ are perpendicular to $BC$. Find measure of angle $HAG$
So I am a IX grader, and I have been stuck on this problem for about 40 mins and am not able to progress further. First I applied the angle bisector theorem $$\frac{BC}{AB} = \frac{CE}{AE}\tag i$$
$$\frac{CB}{AC}=\frac{BF}{AF}\tag{ii}$$
From equations (i) and (ii) I obtained $$\frac {CE.AB}{AE}=\frac {BF.AC}{AF}$$
I also obtained angle $BAC=90$ degrees from angle sum property however after that I am stuck, I tried constructing parallel lines to BC but no avail, I thought to construct a circumcircle as $DA=DB=DC$ in hope to apply some properties but no progress yet. Any thoughtful comments are highly appreciated.
The angle is equal $45^°$. Use that CHFA and BGEA are inscribed in circles.