I need to prove that this line is a tangent to the circle

Question

The problem is this:

Given two different points, $A$ and $B$, take the midpoint between them ($O$) draw the circumference $\Gamma (O,OA)$

Take any point $C$ on $AB$ and draw a line $t$ perpendicular to $AB$ through $C$

Take any point $P$ on $t$ as long as it lies in the interior of $\Gamma$ and draw $AP$ and $BP$ let these lines touch $\Gamma$ on $D$ and $E$ respectively. Take the midpoint $F$ of $D$ and $E$ and let $G$ be the intersection between $t$ and $OF$.

Show that $GE$ is tangent to $\Gamma$

The draw looks like this:

Answer 1

Let $H$ be an orthocenter of triangle $ABP$. Then lines $BD$, $AE$ and $PC$ are concurrent at $H$.

Let $M$ be a midpoint of $PH$. Then $M$ is a circumcenter of quadrilateral $PDHE$, because $\angle HDP=\angle PEH=\frac \pi 2$. Thus $ME=MD$ and since $OE=OD$ we see that $F \in OM$. Thus $M=G$. So

$$\angle OEG = \pi - \angle GOE - \angle EGO = \pi - \frac 12 \angle DOE - \frac 12 \angle EGD = \pi - \angle DAE - \angle EHD = \angle HDA = \frac \pi 2,$$ which means that $GE$ is tangent to $\Gamma$.

Answer 2

By considering $\angle AEP$ and $\angle ACP$, we have $EACP$ is a cyclic quadrilateral.

$∴ z = y$

F is the midpoint of the chord $DE \rightarrow x_1 = x_2$

By considering (angle at center) = 2*(angle at circumference), we have $y = 0.5*(x_1 + x_2) = … = x_1$

Summing up, we have $z = x_1 \rightarrow EOCG$ is a cyclic quadrilateral.

Result follows by applying “opposite angles of a cyclic quadrilateral".