Let $A$ be the altitude from $A$ of a triangle $ABC$. Let $P$ be the intersection of said altitude and the circumference of the triangle. Prove that the simson line of $P$ is parallel to the tangent of the circle through $A$.
I made a segment that goes from $E$ to $G$ such that segment its perpendicular to $GH$, and the same with $FH$. I would like to prove that the angle $FEG = 90^\circ = EFH$. So $EFHG$ its a square for sum of angles and that would prove that $EF$ its parallel to $GH$.
I am using the same letters as above, introduced by the picture rather. It may be that the arguments depend on the picture, so for instance by the fact that $\hat A<90^\circ$, but i hope things can a posteriori be arrange for all cases. It is enough to show that the angle $\widehat{HAD}$ (which is half of the measure of the arc from $A$ through $B$ to $P$ in the picture) is the same as $\widehat ADE$, and here is the line for this using that $PDEC$ is cyclic (two right angles in $D$, $E$): $$ \widehat{ADE} = \widehat{ECP} = \widehat{ACP} = \frac 12(\overset\frown{AB}+\overset\frown{BP}) = \widehat{HAP}\ . $$