Perpendicularity in a given triangle

Question

I was asked to solve the following problem by a friend:

Here, $BC$ is a diameter of the circle, $E$ is the midpoint of the $DC$ arc, $F$ is the midpoint of $BD$, $G$ is the intersection of $FE$ with the circumference, and $H$ is the intersection of the tangents to the circumference passing through $B$ and $D$. Prove that the $\angle HGE$ is right.

I noticed that $\square DEFC$ is inscribed, noticed that $GE$ is a bisector of $\angle CGD$, noticed that $\triangle AFB$ is similar by a ratio of $1:2$ to $\triangle CDB$, but my great problem is to find the relation between $\angle GCD$ and $\angle CBD$. If I find the relation between this angles, I would be able to solve the problem.

Any help will be aprrectiated.

Thanks for attention.

Answer 1

It is easy to prove that $\,\triangle EAF\,$ is a right-angled triangle.

Moreover, it is also easy to prove that $\,\triangle HFB\,$ and $\,\triangle DFA\,$ are similar triangles, hence,

$HF:FB=DF:AF\qquad\color{blue}{(1)}$

On the other hand, by applying the intersecting cords theorem, we get that:

$GF:FB=DF:EF\qquad\color{blue}{(2)}$

From $\,(1)\,$ and $\,(2)\,,\,$ it follows that

$HF:GF=EF:AF$

So the triangles $\,\triangle HGF\,$ and $\,\triangle EAF\,$ are similar because they have two pairs of corresponding proportional sides and the angles $\angle HFG$ and $\angle EFA$ between them are equal.

Since $\,\triangle EAF\,$ is a right-angled triangle, then $\,\triangle HGF\,$ is a right-angled triangle too.

Answer 2

Let $G'$ be the second intersection of the circumcircle of $\triangle BDC$ with the circumcircle of $\triangle AHE$. By radical Axis theorem we conclude that $F$ is the intersection of $HA$, $BD$ and $GE$, implying that $G' = G$. Thus $AGHE$ is cyclic. It suffices to show that $\angle EAH = 90$. Let $P$ be the intersection of $DC$ with $AE$. Then by Thales Theorem $FAPD$ is cyclic. We get $90 = \angle FDP = \angle PAF = \angle EAH$