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I did not get directed angles and did not use the hint. This is my approach:

In the picture circle c$c$ is on points A$A$, E$E$ and F$F$.Circle d Circle $d$ is on points E$E$, F$F$, B$B$ and C$C$. GE$GE$ is tangent to circle d$d$ at E$E$, since ED$ED$ is a diameter of circle d$d$, then we have $GE\bot EM$ hencMEhence $ME$ is tangent to the circle c$c$.Now Now triangles GEM$\triangle GEM$ and GFM$\triangle GFM$ are equalcongruent due to SSS$SSS$.So So we have :

$\angle GFM=\angle GEM=90^o$$$\angle GFM=\angle GEM=90^{\circ}$$

This means that MF$MF$ is also tangent to circle c$c$. Now for line $IJ||BC$ ,$IJ\parallel BC$,H $H$ is the orthocenter of triangle ABC$\triangle ABC$, so AH$AH$ is perpendicular to BC$BC$. AH$AH$ is also the diameter of circle c$c$. since $IJ||BC$$IJ\parallel BC$ then $AH\bot IJ$$AH\perp IJ$, this means that IJ$IJ$ is tangent to circle c$c$ at point A$A$.

enter image description here

I did not get directed angles and did not use the hint. This is my approach:

In the picture circle c is on points A, E and F.Circle d is on points E, F, B and C. GE is tangent to circle d at E, since ED is a diameter of circle d, then we have $GE\bot EM$ hencME is tangent to the circle c.Now triangles GEM and GFM are equal due to SSS.So we have :

$\angle GFM=\angle GEM=90^o$

This means that MF is also tangent to circle c. Now for line $IJ||BC$ ,,H is the orthocenter of triangle ABC, so AH is perpendicular to BC. AH is also the diameter of circle c. since $IJ||BC$ then $AH\bot IJ$, this means that IJ is tangent to circle c at point A.

enter image description here

I did not get directed angles and did not use the hint. This is my approach:

In the picture circle $c$ is on points $A$, $E$ and $F$. Circle $d$ is on points $E$, $F$, $B$ and $C$. $GE$ is tangent to circle $d$ at $E$, since $ED$ is a diameter of circle $d$, then we have $GE\bot EM$ hence $ME$ is tangent to the circle $c$. Now triangles $\triangle GEM$ and $\triangle GFM$ are congruent due to $SSS$. So we have :

$$\angle GFM=\angle GEM=90^{\circ}$$

This means that $MF$ is also tangent to circle $c$. Now for line $IJ\parallel BC$, $H$ is the orthocenter of triangle $\triangle ABC$, so $AH$ is perpendicular to $BC$. $AH$ is also the diameter of circle $c$. since $IJ\parallel BC$ then $AH\perp IJ$, this means that $IJ$ is tangent to circle $c$ at point $A$.

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enter image description here

I did not get directed angles and did not use the hint. This is my approach:

In the picture circle c is on points A, E and F.Circle d is on points E, F, B and C. GE is tangent to circle d at E, since ED is a diameter of circle d, then we have $GE\bot EM$ hencME is tangent to the circle c.Now triangles GEM and GFM are equal due to SSS.So we have :

$\angle GFM=\angle GEM=90^o$

This means that MF is also tangent to circle c. Now for line $IJ||BC$ ,,H is the orthocenter of triangle ABC, so AH is perpendicular to BC. AH is also the diameter of circle c. since $IJ||BC$ then $AH\bot IJ$, this means that IJ is tangent to circle c at point A.